Full text of "Principles of Refrigeration"

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WILEY

TOPPAN

Wiley International Edition

PRINCIPLES OF
REFRIGERATION

ROY J. DOSS AT, Associate Professor of Refrigeration
and Air Conditioning, University of Houston, Houston, Texas

JOHN WILEY & SONS, INC.
NEW YORK and LONDON

TOPPAN COMPANY, LTD.
TOKYO, JAPAN

Authorized reprint of the edition published by John
Wiley & Sons, Inc., New York and London.

All Rights Reserved. No part of this book may
be reproduced in any form without the written
permission of John Wiley & Sons, Inc.

Wiley International Edition

This book is not to be sold outside the country
to which it is consigned by the publisher.

Library of Congress Catalog Card Number: 61-15396

Printed in Singapore by Toppan Printing Co. (S) Pte. Ltd.

Preface

This textbook has been written especially for use in programs where a full
curriculum in refrigeration is offered. However, the material covered and the
method of presentation are such that the text is also suitable for adult evening
classes and for on-the-job training and self-instruction. Furthermore, the
material is so arranged and sectionalized that this textbook is readily adaptable
to any level of study and to any desired method or sequence of presentation.

Despite a rigorous treatment of the thermodynamics of the cycle, application
of the calculus is not required nor is an extensive background in physics and
thermodynamics presupposed. The first four chapters deal with the funda-
mental principles of physics and thermodynamics upon which the refrigeration
cycle is based. For those who are already familiar with these fundamentals,
the chapters will serve as review or reference material.

Chapter 21 treats electric motors and control circuits as they apply to refrigera-
tion and air conditioning systems. This material is presented from the viewpoint
of practical application, the more mathematical approach being left to companion
electrical courses.

Throughout this textbook emphasis is placed on the cyclic nature of the
refrigeration system, and each part of the system is carefully examined in relation
to the whole. Too, care is taken continually to correlate theory and practice
through the use of manufacturer's catalog data and many sample problems. To
this end, certain pertinent catalog data are included.

Roy J. Dossat

July, 1961

Acknowledgments

Most of the material in this textbook
is based on information gathered from
publications of the American Society of
Heating, Refrigerating, and Air Condition-
ing Engineers and of the following equip-
ment manufacturers :

Acme Industries, Inc.
Alco Valve Company
Anaconda Metal Hose Division, The

American Brass Company
Bell & Gossett Company
Carrier Corporation
Controls Company of America
Dean Products, Inc.
Detroit Controls Division, American

Radiator & Standard Sanitary

Corporation
Detroit Ice Machine Company
Dole Refrigerating Company
Dunham-Bush, Inc.
Edwards Engineering Corporation
E. I. du Pont de Nemours & Company
Freezing Equipment Sales, Inc.
Frick Company
General Controls Company
General Electric Company
Halstead & Mitchell

Ingersoll-Rand Company

Kennard Division, American Air Filter

Company, Inc.
Kramer Trenton Company
McQuay, Inc.
The Marley Company
Marsh Instrument Company
Mueller Brass Company
Penn Controls, Inc.
Recold Corporation
Sporlan Valve Company
Tecumseh Products Company
Tranter Manufacturing, Inc.
Tubular Exchanger Manufacturers

Association, Inc.
Tyler Refrigeration Corporation
The Vilter Manufacturing Company
worthington corporation
York Corporation, Subsidiary of Borg-

Warner Corporation

Appreciation is expressed to all these
organizations for their contributions in the
form of photographs and other art work,
and for granting permission to reproduce
proprietary data, without which this text-
book would not have been possible.

Contents

1. Pressure, Work, Power, Energy I

2. Matter, Internal Energy, Heat, Temperature 10

3. Thermodynamic Processes 24

4. Saturated and Superheated Vapors 43

5. Psychrometric Properties of Air 57

6. Refrigeration and the Vapor Compression System 71

7. Cycle Diagrams and the Simple Saturated Cycle 89

8. Actual Refrigerating Cycles 107

9. Survey of Refrigeration Applications 121

10. Cooling Load Calculations 144

11. Evaporators 164

12. Performance of Reciprocating Compressors 203

13. System Equilibrium and Cycling Controls 225

14. Condensers and Cooling Towers 244

15. Fluid Flow, Centrifugal Liquid Pumps, Water and Brine Piping 274

16. Refrigerants 284

17. Refrigerant Flow Controls 298

vii

viii PRINCIPLES OF REFRIGERATION

18. Compressor Construction and Lubrication 334

19. Refrigerant Piping and Accessories 365

20. Defrost Methods, Low-Temperature Systems, and Multiple Temperature
Installations 388

21. Electric Motors and Control Circuits 407
Tables and Charts 430

Index 535

Pressure, Work,
Power, Energy

l-l. Force. A force is defined as a push or a
pull. It is anything that has a tendency to set
a body in motion, to bring a moving body to
rest, or to change the direction of motion. A
force may also change the size or shape of a
body. That is, the body may be twisted, bent,
stretched, compressed, or otherwise distorted
by the action of a force.

The most familiar force is weight. The
weight of a body is a measure of the force
exerted on the body by the gravitational pull
of the earth (Fig. 1-1).

There are many forces other than the force of
gravity, but all forces are measured in weight
units. Although the most commonly used unit
of force measure is the pound, any unit of
weight measure may be used, and the particular
unit- used at any time will usually depend on
the magnitude of the force to be measured.
1-2. Pressure. Pressure is the force exerted
per unit of area. It may be described as a
measure of the intensity of a force at any given
point on the contact surface. Whenever a force
is evenly distributed over a given area, the
pressure at any point on the contact surface is
the same and can be calculated by dividing the
total force exerted by the total area over which
the force is applied. This relationship is
expressed by the following equation

where P «■ the pressure expressed in units of F
per unit of A
F — the total force in any units of force
A = the total area in any units of area
1-3. Measurement of Pressure. As indi-
cated by equation 1-1, pressure is measured
in units of force per unit of area. Pressures are
most frequently given in pounds per square
inch, abbreviated psi. However, pressure, like
force, as a matter of convenience and depending
on the magnitude of the pressure, may be stated
in terms of other units of force and area, such
as pounds per square foot, tons per square foot,
grams per square centimeter, etc.

Example l-l. A rectangular tank, measur-
ing 2 ft by 3 ft at the base, is filled with water.
If the total weight of the water is 432 lb, deter-
mine the pressure exerted by the water on the
bottom of the tank in

(a) pounds per square foot

(b) pounds per square inch

Solution

(a) Area of tank base

Total weight of water

'-h

(1-1)

2x3

= 6sqft

= 432 lb

432
Applying Equation 1-1, P —^

= 72psf
(b) Area of the tank base 24 x 36

= 864 sq in.
Total weight of water = 432 lb

432
Applymg Equation 1-1, P

864
= 0.5 psi

The problem described in Example 1-1 is
illustrated in Fig. 1-2. Notice that the pressure
on the bottom of the tank in pounds per square
foot is equivalent to the downward force
exerted by the weight of a column of water
having a cross section of one square foot, where-
as the pressure in pounds per square inch is
equivalent to the downward force exerted by a
column of water having a cross section of 1 sq in.
Further, since there are 144 sq in. in 1 sq ft, the
force exerted per square foot is 1 44 times as great
as the force exerted per square inch.
1-4. Atmospheric Pressure. The earth is
surrounded by an envelope of atmosphere or air
which extends upward from the surface of the
earth to a distance of some 50 mi or more. Air
has weight and, because of its weight, exerts a

PRINCIPLES OF REFRIGERATION

Pointer -

-Spring scale

Weight

Fig. 1-1, Because of gravity, the suspended weight
exerts a\downward force of 7 lb.

pressure on the surface of the earth. The pressure
exerted by the atmosphere is known as atmos-
pheric pressure.

The weight of a column of air having a cross
section of 1 sq in. and extending from the surface
of the earth at sea level to the upper limits of the
atmosphere is 14.696 lb. Therefore, the pressure
on the surface of the earth at sea level resulting
from the weight of the atmosphere is 14.696 psi
(14.7). This is understood to be the normal or
standard atmospheric pressure at sea level and
is sometimes referred to as a pressure of one
atmosphere. Actually, the pressure of the
atmosphere does not remain constant, but will
usually vary somewhat from hour to hour
depending upon the temperature, water vapor
content, and several other factors.

Because of the difference in the height of the
column, the weight of a column of air of given

cross section will be less when taken at an
altitude of one mile above sea level than when
taken at sea level. Therefore, it follows that
atmospheric pressure decreases as the altitude
increases.

1-5. Barometers. Barometers are instruments
used to measure the pressure of the atmosphere
and are of several types. A simple barometer
which measures atmospheric pressure in terms
of the height of a column of mercury can be
constructed by filling with mercury a hollow
glass tube 36 in. or more long and closed at one
end. The mercury is held in the tube by placing
the index finger over the open end of the tube
while the tube is inverted in an open dish of
mercury. When the finger is removed from the
tube, the level of the mercury in the tube will
fall, leaving an almost perfect vacuum at the
closed end. The pressure exerted downward by
the atmosphere on the open dish of mercury will
cause the mercury to stand up in the evacuated
tube to a height depending upon the amount of
pressure exerted. The height of the mercury
column in the tube is a measure of the pressure
exerted by the atmosphere and is read in inches
of mercury column (abbreviated in. Hg). The
normal pressure of the atmosphere at sea level
(14.696 psi) pressing down on the dish of mercury
will cause the mercury in the tube to rise to a

Fig. 1-2. Of the total weight of the water in the
container, that part which is exerted on a I sq ft area
is the pressure in pounds per square foot. Likewise,
that part which is exerted on a I sq in. area is the
pressure in pounds per square inch.

PRESSURE, WORK, POWER, ENERGY 3

height of 29.921 in. (Fig. 1-3). A column of
mercury 29.921 in. high is, then, a measure of a
pressure equivalent to 14.696 psi. By dividing
29.921 in. Hg by 14.696 psi, it is determined that
a pressure of 1 psi is equivalent to a pressure of
2.036 in. Hg. Therefore, 1 in. Hg equals
1/2.036, or 0.491 psi, and the following equa-
tions are established:

psi

in. Hg = ■

(1-2)

0.491

and psi = in. Hg x 0.491 (1-3)

Example 1-2. What is the pressure of the
atmosphere in psi if a barometer reads 30.2
in. Hg?

Solution. Applying Equation 1-3,
P - 30.2 x 0.491
= 14.83 psi
Example 1-3. In Fig. 1-3, how high will
the mercury stand in the tube when the atmos-
pheric pressure is 14.S psi?

Solution. Applying Equation 1-2,

0.491
= 29.53 in. Hg

1-6. Pressure Gages. Pressure gages are
instruments used to measure the fluid pressure
(either gaseous or liquid) inside a closed vessel.
Pressure gages commonly used in the refriger-
ation industry are of two types: (1) manometer
and (2) bourdon tube.

1-7. Manometers. The manometer type gage
utilizes a column of liquid to measure the
pressure, the height of the column indicating
the magnitude of the pressure. The liquid used
in manometers is usually either water or mercury.
When mercury is used, the instrument is known
as a mercury manometer or mercury gage and,
when water is used, the instrument is a water
manometer or water gage. The simple barom-
eter described previously is a manometer type
instrument.

A simple mercury manometer, illustrated in
Figs. 1-4a, 1-46 and l-4c, consists of a U-shaped
glass tube open at both ends and partially filled
with mercury. When both legs of the U-tube
are open to the atmosphere, atmospheric pres-
sure is exerted on the mercury in both sides of
the tube and the height of the two mercury
columns is the same. The height of the two
mercury columns at this position is marked as
the zero point of the scale and the scale is cali-

brated in inches to read the deviation of the
mercury columns from the zero condition in
either direction (Fig. l-4a).

When in use, one side of the U-tube is
connected to the vessel whose pressure is to be
measured. The pressure in the vessel, acting on
one leg of the tube, is opposed by the atmos-
pheric pressure exerted on the open leg of the
tube. If the pressure in the vessel is greater
than that of the atmosphere, the level of the
mercury on the vessel side of the U-tube is
depressed while the level of the mercury on the
open side of the tube is raised an equal amount
(Fig. l-4b). If the pressure in the vessel is less
than that of the atmosphere, the level of the
mercury in the open leg of the tube is depressed
while the level of the mercury in the leg con-
nected to the vessel is raised by an equal amount
(Fig. l-4c). In either case, the difference in the
heights of the two mercury columns is a
measure of the difference in pressure between
the total pressure of the fluid in the vessel and
the pressure of the atmosphere.

In Fig. l-4b, the level of the mercury is 2 in.

Scale (inches)

Dish of mercury

Fig. 1-3. The pressure exerted by the weight of the
atmosphere on the open dish of mercury causes the
mercury to stand up into the tube. The magnitude
of the pressure determines the height of the mercury
column.

PRINCIPLES OF REFRIGERATION

Atmospheric
/pressure \

Fig. l-4o. Simple U-tube manometer. Since both
legs of the manometer are open to the atmosphere
and are at the same pressure, the level of the mercury
is the same in both sides.

Atmospheric
pressure
30 in. Hg

Fig. I -4b. Simple manometer indicates that the
vessel pressure exceeds the atmospheric pressure

vessel pressure
by 4 in. Hg.

below the zero point in the side of the U-tube
connected to the vessel and 2 in. above the zero
point in the open side of the tube. This indi-
cates that the pressure in the vessel exceeds the
pressure of the atmosphere by 4 in. Hg (1.96
psi). In Fig. l-4c, the level of the mercury is
depressed 2 in. in the side of the tube open to
the atmosphere and raised 2 in. in the side con-
nected to the vessel, indicating that the pressure
in the vessel is 4 in. Hg (1.96 psi) below (less
than) atmospheric. Pressures below atmospheric
are usually called "vacuum" pressures and may
be read as "inches of mercury, vacuum."

Manometers using water as the measuring
fluid are particularly useful for measuring very
small pressures. Because of the difference in the
density of mercury and water, pressures so
slight that they will not visibly affect the height
of a mercury column will produce easily
detectable variations in the height of a water
column. Atmospheric pressure, which will
support a column of mercury only 29.921 in.
high, will lift a column of water to a distance of
approximately 34 ft. A pressure of 1 psi will
raise a column of water 2.31 ft or 27.7 in. and a

Atmospheric
pressure
30 in. Hg

Vessel

- pressure

26 in. Hg

Fig. l-4c Manometer indicates that the vessel
pressure is 4 in. Hg less than the atmospheric
pressure of 30 in. Hg.

Fig. 1-5. Bourdon tube gage mechanism. (Courtesy
Marsh Instrument Company.)

pressure of only 0.036 psi is sufficient to support
a column of water 1 in. high. Hence, 1 in.
of water column is equivalent to 0.036 psi.

Table 1-1 gives the relationship between the
various units of pressure measurement.
1-8. Bourdon Tube Gages. Because of the
excessive length of tube required, gages of the
manometer type are not practical for measuring
pressures above IS psi and are more or less

PRESSURE, WQRK, POWER, ENERGY 5

inches of mercury (Fig. 1-66). In many cases,
single gages, known as "compound" gages, are
designed to measure pressures both above and
below atmospheric (Fig. l-6c). Such gages are
calibrated to read in psi above atmospheric and
in inches of mercury below atmospheric.
1-9. Absolute and Gage Pressures. Absolute
pressure is understood to be the "total" or
"true" pressure of a fluid, whereas gage pressure

<°)

(b)

(c)

Fig. 1-6. Typical bourdon tube gages, (a) Pressure gage, (b) Vacuum gage, (c) Compound gage. (Courtesy
Marsh instrument Company.)

limited to the measurement of relatively small
pressures in air ducts, etc.

Gages of the bourdon tube type are widely
used to measure the higher pressures en-
countered in refrigeration work. The actuating
mechanism of the bourdon tube gage is illus-
trated in Fig. 1-5. The bourdon tube, itself, is a
curved, elliptical-shaped, metallic tube which
tends to straighten as the fluid pressure in the
tube increases and to curl tighter as the pressure
decreases. Any change in the curvature of the
tube is transmitted through a system of gears
to the pointer. The direction and magnitude
of the pointer movement depend on the direc-
tion and magnitude of the change in the curva-
ture of the tube.

Bourdon tube gages are very rugged and will
measure pressures either above or below
atmospheric pressure. Those designed to
measure pressures above atmospheric are known
as "pressure" gages (Fig. l-6a) and are gener-
ally calibrated in psi, whereas those designed to
read pressures below atmospheric are called
"vacuum" gages and are usually calibrated in

is the pressure as indicated by a gage. It is
important to understand that gages are cali-
brated to read zero at atmospheric pressure and
that neither the manometer nor the bourdon
tube gage measures the "total" or "true"
pressure of the fluid in a vessel; both measure
only the difference in pressure between the total
pressure of the fluid in the vessel and the atmos-
pheric pressure. When the fluid pressure is
greater than the atmospheric pressure, the
absolute pressure of the fluid in the vessel is
determined by adding the atmospheric pressure
to the gage pressure, and, when the fluid
pressure is less than atmospheric, the absolute
pressure of the fluid is found by subtracting the
gage pressure from the atmospheric pressure.
The relationship between absolute pressure and
gage pressure is shown graphically is Fig. 1-7.

Example 1-4. A pressure gage on a
refrigerant condenser reads 120 psi. What is
the absolute pressure of the refrigerant in the
condenser?

Solution. Since the barometer reading is not
given, it is assumed that the atmospheric

PRINCIPLES OF REFRIGERATION

Gage
pressure

45-
40
35-
30
25-
20-
15-
10-
5-

Absolute
pressure

Pressures above
atmospheric in psi

Atmospheric pressure

-Pressures below-
- atmospheric in—
in. Hg

59.7
54.7
49.7

-44.7
39.7
34.7
29.7
24.7

•19.7

29.92 in. Hg
(14.7 psi)

5-
10
15-
20

(14.7 psi) 25
29.92 in. Hg~~

Fig. 1-7. Relationship between absolute and gage
pressures.

pressure is normal at sea level, 14.696 psi, and,
since the pressure of the refrigerant is above
atmospheric, the absolute pressure of the
refrigerant is equal to the gage pressure plus
the atmospheric pressure.

Gage pressure in psi =120

Atmospheric pressure in psi = 14.696

Absolute pressure of
refrigerant = 134.696 psi

Example 1-5. A compound gage on the
suction side of a vapor compressor reads
5 in. Hg, whereas a barometer nearby reads
29.6 in. Hg. Determine the absolute pressure
of the vapor entering the compressor.

Solution. Since the pressure of the vapor
entering the compressor is less than atmospheric,
the absolute pressure of the vapor is computed
by subtracting the gage pressure from atmos-
pheric pressure.

Atmospheric pressure in
in. Hg = 29.6

Gage pressure in in. Hg = 5.0

Absolute pressure in
in. Hg = 24.6 in. Hg

Absolute pressure 24.6 x 0.491

= 12.08 psi

Example 1-6. During compression the
pressure of a vapor is increased from 10 in. Hg
gage to 125 psi gage. Calculate the total increase
in pressure in psi.

Solution. Since the pressure increases from
10 in. Hg below atmospheric to 125 psi above
atmospheric, the total increase in pressure is
the sum of the two pressures.

Initial pressure = 10 in. Hg
Initial pressure in psi below 1 x 0.491

atmospheric = 4.91 psi
Final pressure in psi above

atmospheric =125 psi

Total increase in pressure = 129.91 psi
Absolute pressure in psi is abbreviated psia,
whereas gage pressure in psi is abbreviated psig.
1-10. Work. Work is done when a force
acting on a body moves the body through a
distance. The amount of work done is the
product of the force and the distance through
which the force acts. This relationship is shown
by the following equation:

W = F x 1 (1-4)

where F = the force applied in any units of
force
1 = the distance through which the force

acts in any linear unit
W = the work done expressed in units of
force and linear measure

The work done is always expressed in the
same unit terms used to express the magnitude
of the force and the distance. For instance, if
the force is expressed in pounds and the dis-
tance in feet, the work done is expressed in
foot-pounds. The foot-pound is the most
frequently used unit of work measure.

Example U7. A ventilating fan weighing
315 lb is hoisted to the roof of a building 200 ft
above the level of the ground. How much
work is done?

Solution. By applying
Equation 1-4, the weight of
the fan

Distance through which
the fan is hoisted

Work done

= 315 lb

= 200 ft

315 x 200
= 63,000 ft-lb

I -I I. Power. Power is the rate of doing
work. That is, it is the work done divided by the
time required to do the work. The unit of power

PRESSURE, WORK, POWER, ENERGY 7

is the horsepower. One horsepower is defined as
the power required to do work at the rate of
33,000 ft-lb per minute or (33,000/60) 550 ft-lb
per second. The power required in horsepower
may be found by either of the following equa-
tions:

^-TSMTT, (1 " 5)

where Hp = the horsepower

W = the work done in foot-pounds

t = the time in minutes

Hp =

(1-6)

550 x t

where t = the time in seconds

Example 1-8. In Example 1-7, if the time
required to hoist the fan to the roof of the
building is 5 minutes, how much horsepower
is required?

Solution. Total work done =63,000 ft-lb

Time required to do the

work = 5 min

„ . J 63,000

Horsepower required 33,000 x 5

= 0.382 hp
1-12. Energy. In order to do work or to
cause motion of any kind, energy is required.
A body is said to possess energy when it has
the capacity for doing work. Hence, energy is
described as the ability to do work. The
amount of energy required to do a given amount
of work is always equal to the amount of work
done and the amount of energy a body possesses
is equal to the amount of work a body can do
in passing from one condition or position to
another.

Energy may be possessed by a body in either
or both of two basic kinds: (1) kinetic and (2)
potential.

1-13. Kinetic Energy. Kinetic energy is the
energy a body possesses as a result of its motion
or velocity. For instance, a hammer swinging
through an arc, a bullet speeding toward a
target, and the moving parts of machinery all
have kinetic energy by virtue of their motion.
The amount of kinetic energy a body possesses
is a function of its mass and its velocity and may
be determined by the following equation:

M x V 2

where JT — the kinetic energy in foot-pounds
M = the weight of the body in pounds
V = the velocity in feet per second (fps)
g = gravitational constant (32. 1 74 ft/sec 2 )

Example 1-9. An automobile weighing
3500 lb is moving at the rate of 30 mph. What
is its kinetic energy?

Solution. Velocity
in fps V

Applying Equa-
tion 1-7, the kinetic
energy AT

5280 ft/mi x 30 mi
3600sec/hr

= 44 fps

3500 lb x (44 fps)*
2 x 32.174 ft/sec 2

= 105,302 ft-lb

1-14. Potential Energy. Potential energy is
the energy a body possesses because of its
position or configuration. The amount of work
a body can do in passing from a given position
or condition to some reference position or
condition is a measure of the body's potential
energy. For example, the driving head of a pile-
driver has potential energy of position when
raised to some distance above the top of a
piling. If released, the driving head can do the
work of driving the piling. A compressed steel
spring or a stretched rubber band possesses
potential energy of configuration. Both the
steel spring and the rubber band have the
ability to do work because of their tendency to
return to their normal condition.

The potential energy of a body may be evalu-
ated by the following equation:

P =M xZ

(1-8)

where P = the potential energy in foot-pounds
M = the weight of the body in pounds
Z = the vertical distance above some
datum or reference

Example 1-10. Ten thousand gallons of
water are stored in a tank located 250 ft above
the ground. Determine the potential energy of
the water in relation to the ground.

Solution. The
weight of the
water in pounds
M

Applying
Equation 1-8,
the potential
energy P

10,000 gal x 8.33 lb/gal
83,300 lb

83,300 lb x 250 ft
20,825,000 ft-lb

8 PRINCIPLES OF REFRIGERATION

1-15. Energy as Stored Work. Before a
body can possess energy, work must be done on
the body. The work which is done on a body
to give the body its motion, position, or con-
figuration is stored in the body as energy.
Hence, energy is stored work. For instance,
work must be done to stretch therubber band, to
compress the steel spring, or to raise the driving
head of a pile-driver to a position above the
piling. In any case, the potential energy stored
is equal to the work done.

The amount of energy a body possesses can be
ascertained by determining the amount of work
done on the body to give the body its motion,
position, or configuration. For example, assume
that the driving head of a pile-driver weighing
200 lb is raised to a position 6 ft above the top
of a piling. The work done in raising the driving
head is 1200 ft-lb (200 lb x 6 ft). Therefore,
1200 ft-lb of energy are stored in the driving-
head in its raised position and, when released,
neglecting friction, the driving-head will do
1200 ft-lb of work on the piling.
1-16. Total External Energy. The total
external energy of a body is the sum of its
kinetic and potential energies.

Example l-l I. Determine the total external
energy of an airplane weighing 10,000 lb and
flying 6000 ft above the ground at a speed of
300 mph.

Solution. Apply-
ing Equation 1-7,

the kinetic energy 10,000 lb x (440 fps),

K 2 x 32.174 ft/sec*

= 30,086,436 ft-lb

Applying
Equation 1-8, the 10,000 lb x 6000 ft

potential energy P - 60,000,000 ft-lb

Adding, the
total external
energy = 90,086,436 ft-lb

1-17. Law of Conservation of Energy. The

First Law of Thermodynamics states in effect
that the amount of energy is constant. None
can be either created or destroyed. Energy is
expended only in the sense that it is converted
from one form to another.
1-18. Forms of Energy* All energy can be
classified as being of either of the two basic
kinds, kinetic or potential. However, energy
may appear in any one of a number of different
forms, such as mechanical energy, electrical

energy, chemical energy, heat energy, etc., and
is readily converted from one form to another.
Electrical energy, for instance, is converted into
heat energy in an electric toaster, heater, or
range. Electrical energy is converted into
mechanical energy in electric motors, solenoids,
and other electrically operated mechanical
devices. Mechanical energy, chemical energy,
and heat energy are converted into electrical
energy in the generator, battery, and thermo-
couple, respectively. Chemical energy is con-
verted into heat energy in chemical reactions
such as combustion and oxidation. These are
only a few of the countless ways in which the
transformation of energy can and does occur.
There are many fundamental relationships
which exist between the various forms of energy
and their transformation, some of which are of
particular importance in the study of refrigera-
tion and are discussed in detail later.

PROBLEMS

1. The cooling tower on the roof of a building
weighs 1360 lb when filled with water. If the
basin of the tower measures 4 ft by 5 ft, what
is the pressure exerted on the roof

(a) in pounds per square foot? Ans. 68 psf.
(£) in pounds per square inch?

Ans. 0.472 psi.

2. If the atmospheric pressure is normal at sea
level and a gage on an R-12 condenser reads
130 psi, what is the absolute pressure of the
Freon in the condenser in pounds per square
inch? Ans. 144.7 psia.

3. What is the total force exerted on the top- of
a piston if the area of the cylinder bore is 5
sq in. and the pressure of the gas in the cylinder
islSQpsi? Ans. 7501b.

4. A barometer reads 10 in. Hg. What is the
atmospheric pressure in psi? Ans. 4.91 psi.

5. A barometer on the wall reads 29.6 in. Hg
while a gage on the tank of an air compressor
indicates 105 psi. What is the absolute pressure
of the air in the tank in pounds per square foot?

Ans. 119.53 psia.

6. A gage on the suction inlet of a compressor
reads 10 in. Hg. Determine the absolute pres-
sure of the suction vapor in psi. Ans. 9.79 psia.

7. A gage on the suction side of a refrigeration
compressor reads 5 in. Hg. If a gage on the
discharge side of the compressor reads 122 psi,
what is the increase in pressure during the
compression? Ans. 124.46 psi.

PRESSURE, WORK, POWER, ENERGY 9

8. An electric motor weighing 236 lb is hoisted
to the roof of a building in 2 min. If the roof
is 125 ft above the ground,

(a) How much work is done?

Ans. 29,500 ft-lb

(b) Neglecting friction and other losses, what
is tiie horsepower required?

Ans. 0.447 hp

9. Compute the kinetic energy of an automobile
weighing 3000 lb and moving at a speed of
75 mph. Ans. 567,188 ft-lb

10. What is the total external energy of the
automobile in Problem 9 if the automobile is
traveling along a highway 6000 ft above
sea level? Ans. 18,000,000 ft-lb

11. What is the total potential energy of 8000
gal of water confined in a tank and located

a mean distance of 135 ft above the
ground? Ans. 8,996,400 ft-lb

12. Water in a river 800 ft above sea level is
flowing at the rate of 5 mph. Calculate the sum
of the kinetic and potential energies per pound
of water in reference to sea level.

Ans. 800.84 ft-lb

13. A water pump delivers 60 gal per minute
of water to a water tank located 100 ft above
the level of the pump. If water weighs 8.33 lb
per gallon and if the friction of the pipe and
other losses are neglected,

(a) How much work is done?

Ans. 49,980 ft-lb
(6) Compute the horsepower required.

Ans. 1.5 hp.

Matter, Internal
Energy, Heat,
Temperature

2-1. Heat. Heat is a form of energy. This
is evident from the fact that heat can be con-
verted into other forms of energy and that other
forms of energy can be converted into heat.
However, there is some confusion as to exactly
what energy shall be termed heat energy.
Popular usage has made the concept of heat as
internal or molecular energy almost universally
accepted. Because of this, referring to heat as
internal energy is almost unavoidable at times.
On the other hand, from a strictly thermo-
dynamic point of view, heat is denned as energy
in transition from one body to another as a
result of a difference in temperature between the
two bodies. Under this concept, all other
energy transfers occur as work. Both these
concepts of heat will evolve in this and the
following chapters. The term heat will be used
hereafter in this book in either sense.
2-2. Matter and Molecules. Everything in
the universe that has weight or occupies space,
all matter, is composed of molecules. Mole-
cules, in turn, are made up of smaller particles
called atoms and atoms are composed of still
smaller particles known as electrons, protons,
neutrons, etc. The study of atoms and sub-
atomic particles is beyond the scope of this
book and the discussion will be limited for the
most part to the study of molecules and their
behavior.

The molecule is the smallest, stable particle
of matter into which a particular substance can
be subdivided and still retain the identity of
the original substance. For example, a grain
of table salt (NaCl) may be broken down into
individual molecules and each molecule will
be a molecule of salt, the original substance.
However, all molecules are made up of atoms,
so that it is possible to further subdivide a
molecule of salt into its component atoms. But,
a molecule of salt is made up of one atom of
sodium and one atom of chlorine. Hence, if a
molecule of salt is divided into its atoms, the
atoms will not be atoms of salt, the original
substance, but atoms of two entirely different
substances, one of sodium and one of chlorine.

There are some substances whose molecules
are made up of only one kind of atoms. The
molecule of oxygen (0 2 ), for instance, is
composed of two atoms of oxygen. If a mole-
cule of oxygen is divided into its two component
atoms, each atom will be an atom of oxygen,
the original substance, but the atoms of oxygen
will not be stable in this condition. They will
not remain as free and separate atoms of oxygen,
but, if permitted, will either join with atoms or
molecules of another substance to form a new
compound or rejoin each other to form again a
molecule of oxygen.

It is assumed that the molecules that make up
a substance are held together by forces of
mutual attraction known as cohesion. These
forces of attraction that the molecules have for
each other may be likened to the attraction that
exists between unlike electrical charges or
between unlike magnetic poles. However,
despite the mutual attraction that exists between
the molecules and the resulting influence that
each molecule has upon the others, the mole-
cules are not tightly packed together. There is a
certain amount of space between them and they
are relatively free to move about. The mole-
cules are further assumed to be in a state of
rapid and constant vibration or motion, the
rate and extent of the vibration or movement
being determined by the amount of energy
they possess.

2-3. Internal Energy. It has been pre-
viously stated that energy is required to do work
or to cause motion of any kind. Molecules,
like everything else, can move about only if they
possess energy. Hence, a body has internal

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE 1 1

energy as well as external energy. Whereas
a body has external mechanical energy because
of its velocity, position, or configuration in
relation to some reference condition, it also has
internal energy as a result of the velocity,
position, and configuration of the molecules of
the materials which make up the body.

The molecules of any material may possess
energy in both kinds, kinetic and potential.
The total internal energy of a material is the
sum of its internal kinetic and potential energies.
This relationship is shown by the equation

U = K + P (2-1)

where U = the total internal energy
K = the internal kinetic energy
P = the internal potential energy

2-4. Internal Kinetic Energy. Internal kinetic
energy is the energy of molecular motion or
velocity. When heat energy flowing into a
material increases the internal kinetic energy,
the velocity or motion of the molecules is
increased. The increase in molecular velocity is
always accompanied by an increase in the
temperature of the material. Hence, a material's
temperature is, in a sense, a measure of the
average velocity of the molecules which make
up the material. The more kinetic energy the
molecules have, the greater is their movement
and the faster they move. The more rapid the
motion of the molecules, the hotter is the
material and the more internal kinetic energy
the material has. It follows, then, that if the
internal kinetic energy of the material is di-
minished by the removal of heat, the motion of
the molecules will be slowed down or retarded
and the temperature of the material will be
decreased.

According to the kinetic theory if the removal
of heat continues until the internal kinetic
energy of the material is reduced to zero, the
temperature of the material will drop to Abso-
lute Zero (approximately —460° 10 and the
motion of the molecules will cease entirely.*

* It is now known that the energy is not zero at
Absolute Zero. It is the disorganization (entropy)
which diminishes to zero. Heat is sometimes
defined as "disorganized energy." Both the energy
and the disorganization decrease as the temperature
decreases. However, the disorganization decreases
faster than the energy and therefore diminishes
to zero before the energy reaches zero.

2-5. States of Matter. Matter can exist in
three different phases or states of aggregation:
solid, liquid, or a vapor or gas. For example,
water is a liquid, but this same substance can
exist as ice, which is a solid, or as steam, which
is a vapor or gas.

2-6. The Effect of Heat on the State of
Aggregation. Many materials, under the
proper conditions of pressure and temperature,
can exist in any and all of the three physical
states of matter. It will be shown presently
that the amount of energy the molecules of the
material have determines not only the tem-
perature of the material but also which of the
three physical states the material will assume at
any particular time. In other words, the addi-
tion or removal of heat can bring about a change
in the physical state of the material as well as a
change in its temperature.

That heat can bring about a change in the
physical state of a material is evident from the
fact that many materials, such as metals will
become molten when sufficient heat is applied.
Furthermore, the phenomenon of melting ice
and boiling water is familiar to everyone. Each
of these changes in the physical state is brought
about by the addition of heat.
2-7. Internal Potential Energy. Internal
potential energy is the energy of molecular
separation or configuration. It is the energy the
molecules have as a result of their position in
relation to one another. The greater the degree
of molecular separation, the greater is the inter-
nal potential energy.

When a material expands or changes its
physical state with the addition of energy, a
rearrangement of the molecules takes place
which increases the distance between them.
Inasmuch as the molecules are attracted to one
another by forces which tend to pull them to-
gether, internal work must be done in order to
separate further the molecules against their
attractive forces. An amount of energy equal to
the amount of internal work done must flow
into the material. This energy is set up in the
material as an increase in the internal potential
energy. It is "stored" energy which is
accounted for by the increase in the mean dis-
tance between the molecules. The source of
this energy is the heat energy supplied.

It is important to understand that in this
instance the energy flowing into the material

12 PRINCIPLES OF REFRIGERATION

has no effect on molecular velocity (internal
kinetic energy); only the degree of molecular
separation (the internal potential energy) is
affected.

2-8. The Solid State. A material in the solid
state has a relatively small amount of internal
potential energy. The molecules of the material
are rather closely bound together by each other's
attractive forces and by the force of gravity.
Hence, a material in the solid state has a rather
rigid molecular structure in which the position
of each molecule is more or less fixed and the
motion of the molecules is limited to a vibratory
type of movement which, depending upon the
amount of internal kinetic energy the molecules
possess, may be either slow or rapid.

Because of its rigid molecular structure, a
solid tends to retain both its size and its shape.
A solid is not compressible and will offer
considerable resistance to any effort to change
its shape.

M. The Liquid State. The molecules of a
material in the liquid state have more energy
than those of a material in the solid state and
they are not so closely bound together. Their
greater energy allows them to overcome each
other's attractive forces to some extent and to
have more freedom to move about. They are
free to move over and about one another in
such a way that the material is said to "flow."
Although a liquid is noncompressible and will
retain its size, because of its fluid molecular
structure, it will not retain its shape, but
will assume the shape of any containing
vessel.

2-10. The Vapor or Gaseous State. The
molecules of a material in the gaseous state have
an even greater amount of energy than those
of a material in the liquid state. They have
sufficient energy to overcome all restraining
forces. They are no longer bound by each
other's attractive forces, neither are they bound
by the force of gravity. Consequently, they
fly about at high velocities, continually collid-
ing with each other and with the walls of the
container. For this reason, a gas will retain
neither its size nor its shape. It is readily com-
pressible and will completely fill any container
regardless of size. Further, if the gas is not
stored in a sealed container, it will escape from
the container and be diffused into the surround-
ing air.

2-11. Temperature. Temperature is a prop-
erty of matter. It is a measure of the level of
heat intensity or the thermal pressure of a body.
A high temperature indicates a high level of heat
intensity or thermal pressure, and the body is
said to be hot. Likewise, a low temperature
indicates a low level of heat intensity or thermal
pressure and the body is said to be cold.
2-12. Thermometers. The most frequently
used instrument for measuring temperature is
the thermometer. The operation of most ther-
mometers depends upon the property of a liquid
to expand or contract as its temperature is
increased or decreased, respectively. Because
of their low freezing temperatures and relatively
constant coefficients of expansion, alcohol and
mercury are the liquids most frequently used in
thermometers. The mercury thermometer is the
more accurate of the two because its coefficient
of expansion is more constant through a greater
temperature range than is that of alcohol.
However, mercury thermometers have the dis-
advantage of being more expensive and more
difficult to read. Alcohol is cheaper and can be
colored for easy visibility.

Two temperature scales are in common use
today. The Fahrenheit scale is used in English
speaking countries, whereas the Centigrade
scale is widely used in European countries as
well as for scientific purposes.
2-13. Centigrade Scale. The point at which
water freezes under atmospheric pressure is
taken as the arbitrary zero point on the Centi-
grade scale, and the point at which water boils
is designated as 100. The distance on the scale
between these two points is divided into one
hundred equal units called degrees, so that the
distance between the freezing and boiling points
of water on the Centigrade scale is 100°. Water
freezes at 0° Centigrade and boils at 100° Centi-
grade.

2-14. Fahrenheit Scale. Although there is
some disagreement as to the actual method used
by Fahrenheit in designing the first temperature
scale, it was arrived at by means similar to those
described in the previous section. On the
Fahrenheit scale, the point at which water
freezes is marked as 32, and the point at which
water boils 212. Thus, there are 180 units
between the freezing and boiling points of
water. The zero or reference point on the
Fahrenheit scale is placed 32 units or degrees

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE 13

below the freezing point of water and is assumed
to represent the lowest temperature Fahrenheit
could achieve with a mixture of ammonium
chloride and snow.

2-15. Temperature Conversion. Temper-
ature readings on one scale can be converted to
reading on the other scale by using the appro-
priate of the following equations:

° F = 9/5° C + 32 (2-2)

° C = 5/9(° F - 32) (2-3)

It should be noted that the difference between
the freezing and boiling points of water on the
Fahrenheit scale is 180°, whereas the difference
between these two points on the Centigrade
scale is only 100°. Therefore, 100 Centigrade
degrees are equivalent to 180 Fahrenheit
degrees. This establishes a relationship such
that 1° C equals 9/5° F (1.8° F) and 1° F equals
5/9° C (0.555° C). This is shown graphically
in Fig. 2-1. Since 0° on the Fahrenheit scale is
32° F below the freezing point of water, it is
necessary to add 32° F to the Fahrenheit equiva-
lent after converting from Centigrade. Like-
wise, it is necessary to subtract 32° F from a
Fahrenheit reading before converting to Centi-
grade.

Example 2-1. Convert a temperature read-
ing of 50° C to the equivalent Fahrenheit
temperature.

Solution. Applying
Equation 2-2, ° F

212*

9/5(50° Q + 32
= 122° F

Example 2-2. A thermometer on the wall
of a room reads 86° F. What is the room
temperature in degrees Centigrade?

Solution. Applying Formula
2-3, the room temperature 5/9(86-32)

in ° C = 30° C

Example 2-3. A thermometer indicates
that the temperature of a certain quantity of
water is increased 45° F by the addition of
heat. Compute the temperature rise in Centi-
grade degrees.

Solution. Temperature rise in ° F

= 45°F
Temperature rise in ° C 5/9(45° F)

= 25°C

2-16. Absolute Temperature. Tempera-
ture readings taken from either the Fahrenheit

32*
0*

-40"

-460*

Boiling point of water

Freezing point of water

Scales coincide

Absolute zero

100*

0*
-17.8*

-40*

-273'

Fig. 2-1. Comparison of Fahrenheit and Centigrade
temperature scales.

or Centigrade scales are in respect to arbitrarily
selected zero points which, as has been shown,
are not even die same for the two scales. When
it is desired to know only the change in tem-
perature that occurs during a process or the
temperature of a substance in relation to some
known reference point, such readings are
entirely adequate. However, when temperature
readings are to be applied in equations dealing
with certain fundamental laws, it is necessary
to use temperature readings whose reference
point is the true or absolute zero of tempera-
ture. Experiment has indicated that such a
point, known as Absolute Zero, exists at approxi-
mately -460° F or -273° C (Fig. 2-1).

Temperature readings in reference to Abso-
lute Zero are designated as absolute tempera-
tures and may be in either Fahrenheit or
Centigrade degrees. A temperature reading on
the Fahrenheit scale can be converted to
absolute temperature by adding 460° to the
Fahrenheit reading. The resulting temperature
is in degrees Rankine (° R).

Likewise, Centigrade temperatures can be
converted to absolute temperatures by adding
273° to the Centigrade reading. The resulting
temperature is stated in degrees Kelvin (° K).

14 PRINCIPLES OF REFRIGERATION

In converting to and from absolute tempera-
tures, the following equations will apply:

(2-4)

T= t+460
t = r-460
T « / + 273

/ - T - 273

(2-5)
(2-6)
(2-7)

where T = absolute temperature in degrees

Rankine or Kelvin
t — temperature in degrees Fahrenheit

or Centigrade
Equations 2-4 and 2-5 apply to the Rankine and
Fahrenheit scales, whereas Equations 2-6 and
2-7 apply to the Kelvin and Centigrade scales.
Hereafter in this book Rankine and Fahrenheit
temperatures are used unless otherwise specified.

Example 2-4. A thermometer on the tank
of an air compressor indicates that the tempera-
ture of the air in the tank is 95° F. Determine
the absolute temperature in degrees Rankine.

Solution. Applying
Equation 2-4, T - 95° F + 460°

= 555° R

Example 2-5. The temperature of the
vapor entering the suction of a refrigeration
compressor is -20° F. Compute the tempera-
ture of the vapor in degrees Rankine.

Solution. Applying
Equation 2-4, T = -20° F + 460°

= 440°R

Example 24. If the temperature of a gas
is 100° C, what is its temperature in degrees
Kelvin?

Solution. Applying
Equation 2-6,

T = 100° C + 273°
= 373° K

Example 2-7. The temperature of steam
leaving a boiler is 610° R. What is the tempera-
ture of the steam on the Fahrenheit scale?

Solution. Applying
Equation 2-5, / = 610° R - 460°

= 150° F

2-17. Direction and Rate of Heat Flow.

Heat will flow from one body to another when,
and only when, a difference in temperature
exists between the two bodies. If the tempera-

ture of the two bodies is the same, there is no
transfer of heat.

Heat always flows down the temperature
scale from a high temperature to a low tem-
perature, from a hot body to a cold body, and
never in the opposite direction. Since heat is
energy and cannot be destroyed, if heat is to
leave one body of material, it must flow into
and be absorbed by another body of material
whose temperature is below that of the body
being cooled.

The rate of heat transfer between two bodies
is always directly proportional to the difference
in temperature between the two bodies.
2-18. Methods of Heat Transfer. The trans-
fer of heat from one place to another occurs
in three ways: (1) conduction, (2) convection,
and (3) radiation.

2-19. Conduction. Heat transfer by con-
duction occurs when energy is transmitted by
direct contact between the molecules of a single
body or between the molecules of two or more
bodies in good thermal contact with each other.
In either case, the heated molecules communi-
cate their energy to the other molecules im-
mediately adjacent to them. The transfer of
energy from molecule to molecule by conduction
is similar to that which takes place between the
balls on a billiard table, wherein all or some
part of the energy of motion of one ball is trans-
mitted at the moment of impact to the other
balls that are struck.

When one end of a metal rod is heated over a
flame, some of the heat energy from the heated
end of the rod will flow by conduction from
molecule to molecule through the rod to the
cooler end. As the molecules at the heated end
of the rod absorb energy from the flame, their
energy increases and they move faster and
through a greater distance. The increased
energy of the heated molecules causes them to
strike against the molecules immediately ad-
jacent to them. At the time of impact and
because of it, the faster moving molecules trans-
mit some of their energy to their slower moving
neighbors so that they too begin to move more
rapidly. In this manner, energy passes from
molecule to molecule from the heated end of the
rod to the cooler end. However, in no case
would it be possible for the molecules furthest
from the heat source to have more energy than
those at the heated end.

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE IS

As heat passes through the metal rod, the air
immediately surrounding the rod is also heated
by conduction. The rapidly vibrating particles
of the heated rod strike against the molecules
of the air which are in contact with the rod.
The energy so imparted to the air molecules
causes them to move about at a higher rate and
communicate their energy to other nearby air
molecules. Thus, some of the heat supplied to
the metal rod is conducted to and carried away
by the surrounding air.

If the heat supply to the rod is interrupted,
heat will continue to be carried away from the
rod by the air surrounding until the tempera-
ture of the rod drops to that of the air. When
this occurs, there will be no temperature differ-
ential, the system will be in equilibrium, and
no heat will be transferred.

The rate of heat transfer by conduction, as
previously stated, is in direct proportion to the
difference in temperature between the high and
low temperature parts. However, all materials
do not conduct heat at the same rate. Some
materials, such as metals, conduct heat very
readily, whereas others, such as glass, wood, and
cork, offer considerable resistance to the con-
duction of heat. Therefore, for any given tem-
perature difference, the rate of heat flow by
conduction through different materials of the
same length and cross section will vary with the
particular ability of the various materials to
conduct heat. The relative capacity of a material
to conduct heat is known as its conductivity.
Materials which are good conductors of heat
have a high conductivity, whereas materials
which are poor conductors have a low con-
ductivity and are used as heat insulators.

In general, solids are better conductors of
heat than liquids, and liquids are better con-
ductors than gases. This is accounted for by
the difference in the molecular structure. Since
the molecules of a gas are widely separated, the
transfer of heat by conduction, that is, from
molecule to molecule, is difficult.
2-20. Convection. Heat transfer by con-
vection occurs when heat moves from one place
to another by means of currents which are set
up within some fluid medium. These currents
are known as convection currents and result
from the change in density which is brought
about by the expansion of the heated portion of
the fluid.

Cooler portions of water descend to
replace the lighter portions that rise

gas^fgje^* - ^^

Flame

Heat is conducted
from flame to
water through
bottom of vessel

Heated portions of water become
lighter and rise toward surface,
thereby distributing the heat
throughout the entire mass

Fig. 2-2. Convection currents set up in a vessel of
water when the vessel is heated at bottom center.

When any portion of a fluid is heated, it ex-
pands and its volume per unit of weight
increases. Thus, the heated portion becomes
lighter, rises to the top, and is immediately
replaced by a cooler, heavier portion of the
fluid. For example, assume that a tank of
water is heated on the bottom at the center
(Fig. 2-2). The heat from the flame is conducted
through the metal bottom of the tank to the
water inside. As the water adjacent to the heat
source absorbs heat, its temperature increases
and it expands. The heated portion of the
water, being lighter than the water surrounding,
rises to the top and is replaced by cooler, more
dense water pushing in from the sides. As this
new portion of water becomes heated, it too
rises to the top and is replaced by cooler water
from the sides. As this sequence continues, the
heat is distributed throughout the entire mass
of the water by means of the convection currents
established within the mass.

Warm air currents, such as those which occur
over stoves and other hot bodies, are familiar to
everyone. How convection currents are utilized
to carry heat to all parts of a heated space is
illustrated in Fig. 2-3.

2-21. Radiation. Heat transfer by radiation
occurs in the form of a wave motion similar to
light waves wherein the energy is transmitted
from one body to another without the need for

16 PRINCIPLES OF REFRIGERATION

Steam coils'
Fig. 2-3. Room heated by natural convection.

intervening matter. Heat energy transmitted
by wave motion is called radiant energy.

It is assumed that the molecules of a body are
in rapid vibration and that this vibration sets
up a wave motion in the ether surrounding the
body.* Thus, the internal molecular energy of
the body is converted into radiant energy waves.
When these energy waves are intercepted by
another body of matter, they are absorbed by
that body and are converted into its internal
molecular energy.

The earth receives heat from the sun by
radiation. The energy of the sun's molecular
vibration is imparted in the form of radiant
energy waves to the ether of interstellar space sur-
rounding the sun. The energy waves travel across
billions of miles of space and impress their
energy upon the earth and upon any other
material bodies which intercept their path. The
radiant energy is absorbed and transformed
into internal molecular energy, so that the
vibratory motion of the hot body (the sun) is
reproduced in the cooler body (the earth).

All materials give off and absorb heat in the
form of radiant energy. Any time the tempera-
ture of a body is greater than that of its sur-
roundings, it will give off more heat by radiation
than it absorbs. Therefore, it loses energy to
its surroundings and its internal energy de-
creases. If the temperature of the body is below
that of its surroundings, it absorbs more radiant
energy than it loses and its internal energy
increases. When no temperature difference

* Ether is the name given to that which fills all
space unoccupied by matter, such as interstellar
space and the space between the molecules of every
material.

exists, the energy exchange is in equilibrium and
the body neither gains nor loses energy.

Heat transfer through a vacuum is impossible
by either conduction or convection, since these
processes by their very nature require that
matter be the transmitting media. Radiant
energy, on the other hand, is not dependent
upon matter as a medium of transfer and there-
fore can be transmitted through a vacuum.
Furthermore, when radiant energy is transferred
from a hot body to a cold body through some
intervening media such as air the temperature of
the intervening media is unaffected by the
passage of the radiant energy. For example,
heat is radiated from a "warm" wall to a "cold"
wall through the intervening air without having
any appreciable effect upon the temperature of
the air. Since the molecules of the air are rela-
tively few and widely separated, the waves of
radiant energy can easily pass between them so
that only a very small part of the radiant energy
is intercepted and absorbed by the molecules of
the air. By far the greater portion of the radiant
energy impinges upon and is absorbed by the
solid wall whose molecular structure is much
more compact and substantial.

Heat waves are very similar to light waves,
differing from them only in length and frequency.
Light waves are radiant energy waves of such
length as to be visible to the human eye. Thus,
light waves are visible heat waves. Whether
heat waves are visible or invisible depends upon
the temperature of the radiating body. For
example, when metal is heated to a sufficiently
high temperature, it will "glow," that is, emit
visible heat waves (light).

When radiant energy waves, either visible or
invisible, strike a material body, they may be
reflected, refracted, or absorbed by it, or they
may pass through it to some other substance
beyond.

The amount of radiant energy which will pass
through a material depends upon the degree of
transparency. A highly transparent material,
such as clear glass or air, will allow most of the
radiant energy to pass through to the materials
beyond, whereas opaque materials, such as
wood, metal, cork, etc., cannot be penetrated
by radiant energy waves and none will pass
through.

The amount of radiant energy which is either
reflected or absorbed by a material depends

MATTER, INTERNAL ENERGY, HEAT. TEMPERATURE 17

upon the nature of the material's surface, that
is, its texture and its color. Materials having a
light-colored, highly polished surface, such as a
mirror, reflect a maximum of radiant energy,
whereas materials having rough, dull, dark
surfaces will absorb the maximum amount of
radiant energy.

2-22. British Thermal Unit. It has already
been established that a thermometer measures
only the intensity of heat and not the quantity.
However, in working with heat it is often
necessary to determine heat quantities. Obvi-
ously, some unit of heat measure is required.

Heat is a form of energy, and as such is
intangible and cannot be measured directly.
Heat can be measured only by measuring the
effects it has on a material, such as the change in
temperature, state, color, size, etc.

The most universally used unit of heat measure
is the British thermal unit, abbreviated Btu. A
Btu is denned as the quantity of heat required to
change the temperature of 1 lb of water 1° F.
This quantity of heat, if added to 1 lb of water,
will raise the temperature of the water 1°F.
Likewise, if 1 Btu is removed from 1 lb of water,
the temperature of the water will be lowered
1°F.

The quantity of heat required to change the
temperature of 1 lb of water 1° F is not a con-
stant amount. It varies slightly with the tem-
perature range at which the change occurs. For
this reason, a Btu is more accurately defined as
being l/180th of the quantity of heat required to
raise the temperature of 1 lb of water from the
freezing point (32° F) to the boiling point
(212° F). This is identified as the "mean Btu"
and is the exact amount of heat required to raise
the temperature of 1 lb of water from 62 to
63° F. If the change in temperature occurs at
any other point on the temperature scale, the
amount of heat involved is either more or less
than the mean Btu, depending upon the par-
ticular point on the temperature scale that the
change takes place. However, the variation
from the mean Btu is so slight that it may be
neglected and, regardless of the temperature
range, for all practical purposes it is sufficiently
accurate to assume that the temperature of 1 lb
of water is changed 1° F by the addition or
removal of 1 Btu.

2-23. Specific Heat. The specific heat of a
material is the quantity of heat required to

change the temperature of 1 pound of the
material 1° F. For instance, the specific heat of
aluminum is 0.226 Btu/lb/°F, whereas that of
brass is 0.089 Btu/lb/°F. This means that 0.226
Btu is required to raise the temperature of 1
pound of aluminum 1° F, whereas only 0.089
Btu is necessary to change the temperature of
1 pound of brass 1° F. Note that by the defini-
tion of the Btu the specific heat of water is 1
Btu per pound per degree Fahrenheit.

The specific heat of any material, like that of
water, varies somewhat throughout the tem-
perature scale. Here again, the variation is so
slight that it is sufficiently accurate for most
calculations to consider the specific heat to be
a constant amount. This is not true, however,
as the material passes through a change in
physical state. The specific heat of a material
in the solid state is approximately one-half that
of the same -material in the liquid state. For
instance, the specific heat of ice is 0.S Btu,
whereas that of water is one. The specific heat
values of materials in the gaseous state are dis-
cussed in another chapter.
2-24. Calculating Heat Quantity. The
quantity of heat which must be added to or
removed from any given mass of material in
order to bring about a specified change in its
temperature can be computed by using the
following equation:

Q, - MC(t t - t0

(2-8)

where Q, = the quantity of heat either absorbed
or rejected by the material
M = the weight of the material in

pounds
C = the specific heat of the material
?! = the initial temperature
t t = the final temperature

Example 2-8. Twenty pounds of water at
an initial temperature of 76° F are heated until
the temperature is increased to 180° F. How
much heat must be supplied?

Solution.
Applying
Equation 2-8,

Q, - 20 lb x 1 x (180 - 76)
= 2080 Btu

Example 2-9. If water weighs 8.33 lb per
gallon, how much heat is rejected by 30 gal of
water in cooling from 80° F to 35° F?

18 PRINCIPLES OF REFRIGERATION

Solution.
Weight of

water in 30 gal x 8.33 lb/gal

pounds = 250 lb

Applying Q, = 250 lb x 1 x (35 - 80)
Equation 2-8, = 250 lb x 1 x (-45)

= 11,250 Btu

Note: Since the specific heat of a material is
given in terms of Btu/lb/° F, the weight of the
material must be determined before Equation
2-8 can be applied.

Where t t is less than t l7 the answer obtained
by applying Equation 2-8 will be negative,
indicating that heat is rejected by rather than
absorbed by the material. In this type of
problem, where the direction of heat flow is
obvious, the negative sign can be ignored and
the answer assumed to be positive.

Example 2-10. Fifteen pounds of cast iron
are cooled from 500° F to 250° F by being im-
mersed in 3 gallons (25 lb) of water whose
initial temperature is 78° F. Assuming that the
specific heat of the cast iron is 0.101 Btu/lb/° F
and that all of the heat given up by the cast iron
is absorbed by the water, what is the final
temperature of the water?

Solution. By
applying Equation
2-8 to compute the
total quantity of heat
given up by the cast
iron,

By rearranging and
applying Equation 2-8
to determine the final
temperature of the
water after absorbing
the heat given up by
the cast iron,

2-25. Heat Divided

Q, = 15 lb x 0.101
x 250

= 378.75 Btu

h MC
378.75

+ 'i

+ 78°F

into

25 x 1
= 15.15° +78
= 93°F
Two Kinds

Categories. It has been previously stated
(Section 2-6) that heat has the ability to bring
about a change in the physical state of a material
as well as the ability to cause a change in its
temperature. Heat is divided into two kinds or
categories, depending upon which of these two
effects it has on a material which either absorbs
or rejects it. The division of heat into several
classifications is made only to facilitate and
simplify certain necessary calculations and does
not stem from any difference in the nature of
heat itself.

2-26. Sensible Heat. When heat either ab-
sorbed or rejected by a material causes or

accompanies a change in the temperature of the
material, the heat transferred is identified as
sensible heat. The term sensible is applied to
this particular heat because the change in tem-
perature it causes can be detected with the sense
of touch and can, of course, be measured with a
thermometer.

2-27. Latent Heat. When heat, either added
to or rejected by a material, brings about or
accompanies a change in the physical state of the
material, the heat is known as latent heat. The
name latent, a Latin word meaning hidden, is
said to have been given to this special kind of
heat by Dr. Joseph Black because it apparently
disappeared into a material without having any
effect on the temperature of the material.

Many materials progressing up the tempera-
ture scale will pass through two changes in the
state of aggregation: first, from the solid to the
liquid phase and then, as the temperature of
the liquid is further increased to a certain level
beyond which it cannot exist as a liquid, the
liquid will change into the vapor state. When
the change occurs in either direction between
the solid and liquid phases, the heat involved is
known as the latent heat of fusion. When the
change occurs between the liquid and vapor
phases, the heat involved is the latent heat of
vaporization.

2-28. Sensible Heat of a Solid. To obtain a
better understanding of the concept of molecular
energy, consider the progressive effects of heat
as it is taken in by a material whose initial
thermodynamic condition is such that its energy
content is zero. Assume that a solid in an open
container is at a temperature of — 460° F
(Absolute Zero). Theoretically, at this tem-
perature the molecules of the material have no
energy and are completely at rest.

When heat energy flows into the solid, the
molecules Of the solid begin to move slowly and
the temperature of the solid begins to climb.
The more heat energy taken in by the solid, the
faster the molecules vibrate and the warmer the
solid becomes. The increase in molecular velo-
city and in the temperature of the solid continues
as more heat is absorbed, until the solid reaches
its melting or fusion temperature. The total
quantity of heat energy required to bring the
temperature of the solid from the original con-
dition of Absolute Zero to the melting or fusion
temperature is known as the sensible heat of the

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE 19

solid. As previously shown, the quantity of heat
which must be transferred in order to bring
about a specified change in the temperature of
any given mass of any material can be calculated
by applying Equation 2-8.
2-29. The Melting or Fusion Temperature.
Upon reaching the fusion temperature, the
molecules of the solid are moving as rapidly as
is possible within the rigid molecular structure
of the solid state. It is not possible to increase
further the motion of the molecules or the tem-
perature of the solid beyond this point without
first overcoming partially the forces of mutual
attraction which exists between the molecules.
Hence, the material cannot exist in the solid
state at any temperature above its melting or
fusion temperature. On reaching the fusion
temperature, any additional heat absorbed by
the material will cause some part of the solid
to revert to the liquid phase.

The exact temperature at which melting or
fusion occurs varies with the different materials
and with the pressure. For instance, at normal
atmospheric pressure, the fusion temperature of
lead is approximately 600° F, whereas copper
melts at approximately 2000° F and ice at only
32° F. In general, the melting temperature
decreases as the pressure increases except for
noncrystalline solids, whose melting tempera-
tures increase as the pressure increases.
2-30. Latent Heat of Fusion. When heat
is absorbed by a solid at the fusion temperature,
the molecules of the solid utilize the energy to
overcome partially their attraction for one
another. They break away from one another to
some extent and become more widely separated.
As the molecules flow over and about one
another, the material loses the rigidity of the
solid state and becomes fluid. It can no longer
support itself independently and will assume the
shape of any containing vessel.

The attraction which exists between the mole-
cules of a solid is considerable and a relatively
large quantity of energy is required to overcome
that attraction. The quantity of heat required
to melt one pound of a material from the solid
phase into the liquid phase is called the latent
heat of fusion. The latent heat of fusion, along
with other values such as specific heat, fusion
temperature, etc., for the different materials has
been determined by experiment and may be
found in various tables.

It is important at this point to emphasize that
the change of phase occurs in either direction at
the fusion temperature, that is, the temperature
at which the solid will melt into the liquid phase
is the same as that at which the liquid will
freeze into the solid phase. Further, the quan-
tity of heat that must be rejected by a certain
weight of liquid at the fusion temperature in
order to freeze into the solid state is exactly
equal to the amount of heat that must "be ab-
sorbed by the same weight of the solid in melting
into the liquid state.

None of the heat absorbed or rejected during
the change of phase has any effect on molecular
velocity. Therefore the temperature of the
material remains constant during the phase
change, and the temperature of the resulting
liquid or solid is the same as the fusion tem-
perature.*

The quantity of heat that is absorbed by a
given weight of a solid at the fusion temperature
in melting into the liquid phase, or, conversely,
the quantity of heat that is rejected by a given
weight of liquid at the fusion temperature in
freezing or solidifying, can be determined by
applying the following equation:

Q L = M x h it (2-9)

where Ql = the quantity of heat in Btu

M = the mass or weight in pounds
h it = the latent heat in Btu per pound

Example 2-11. Calculate the quantity of
heat required to melt 12 lb of ice at 32° F into
water at 32° F. The latent heat of fusion of
water under atmospheric pressure is 144 Btu
per pound.

Solution. Apply-
ing Equation 2-9, the
quantity of heat re-
quired to melt 12 lb
of ice

12 lb x 144 Btu/lb
= 1728 Btu

Note. Since 12 lb of ice absorb 1728 Btu in
melting into water, it follows that 12 lb of water
at 32° F will reject 1728 Btu in returning to the
solid state.

* This applies with absolute accuracy only to
crystalline solids. Noncrystalline solids, such as
glass, have indefinite fusion temperatures. That
is, the temperature will vary during the change of
phase. However, for the purpose of calculating heat
quantities, the temperature is assumed to remain
constant during the phase change.

20 PRINCIPLES OF REFRIGERATION

Example 2-12. If SO lb of ice at 32° F
absorb 6000 Btu, what part of the ice will be
melted?

Solution. By rearranging = _G
and applying Equation 2-9, h if

the part of the ice melted, M _ 6000 Btu

~ 144 Btu/lb
- 41.66 lb

2-31. Sensible Heat of the Liquid. When a
material passes from the solid to the liquid
phase, the resulting liquid is at the fusion tem-
perature. The temperature of the liquid may
then be increased by the addition of heat. Any
heat absorbed by a liquid after the change of
state is set up in the liquid as an increase in the
internal kinetic energy. Molecular velocity
increases and the temperature of the liquid rises.
But here again, as in the case of the solid, the
temperature of the liquid eventually reaches a
point beyond which it cannot be further in-
creased. A liquid cannot exist as a liquid at any
temperature above its vaporizing temperature
for a given pressure and, upon reaching the
vaporizing temperature, if additional heat is
taken in by the liquid, some part of the liquid
will change to the vapor phase.

The total quantity of heat taken in by a liquid
as its temperature is increased from the fusion
to the vaporizing temperature is called the
sensible heat of the liquid. Here again, Equation
2-8, sometimes known as the "sensible heat
equation," can be applied to determine the
quantity of heat necessary to change the tem-
perature of any given weight of liquid through
any specified temperature range.
2-32. Saturation Temperature. The tem-
perature at which a liquid will change into the
vapor phase is called the saturation temperature,
sometimes referred to as the "boiling point"
or "boiling temperature." A liquid whose tem-
perature has been raised to the saturation tem-
perature is called a saturated liquid.

The saturation temperature, that is, the tern'
perature at which vaporization occurs, is
different for each liquid. Iron, for example,
vaporizes at 4450° F, copper at 4250° F, and
lead 3000° F. Water, of course, boils at 212° F,
and alcohol at 170° F. Some liquids boil at
extremely low temperatures. A few of these
are ammonia, oxygen, and helium, which boil
at temperatures of -28° F, -295° F, and
—452° F, respectively.

2-33. Latent Heat of Vaporization. Any

heat taken in by a liquid after the liquid reaches
the saturation temperature is utilized to increase
the degree of molecular separation (increases the
internal potential energy) and the substance
changes from the liquid to the vapor phase.*
There is no increase in molecular velocity and,
therefore, no change in the -internal kinetic
energy during the change in phase. Hence,
the temperature remains constant during the
phase change and the vapor which results is at
the vaporizing temperature.

As the material changes state from a liquid
to a vapor, the molecules of the material acquire
sufficient energy to overcome all restraining
forces, including the force of gravity. The
amount of energy required to do the internal
work necessary to overcome these restraining
forces is very great. For this reason, the capacity
of a material to absorb heat while undergoing
a change from the liquid to the vapor phase is
enormous, many times greater even than its
capacity to absorb heat in changing from the
solid to the liquid phase.

The quantity of heat which 1 lb of a liquid
absorbs while changing into the vapor state is
known as the latent heat of vaporization. The
latent heat of vaporization, like the saturation
temperature, is different for each material. It
will be shown later that both the latent heat
value and the saturation temperature of any
particular liquid vary with the pressure over the
liquid. When the pressure increases, the satur-
ation temperature increases and the latent heat
value decreases.

The quantity of heat required to vaporize any
given weight of liquid at the saturation tem-
perature is calculated by the following equation:

Q L =M xh

(2-10)

where Ql = the quantity of heat in Btu
M = the mass or weight -in pounds
h fg = the latent heat of vaporization in
Btu/lb

Example 2-13. If the latent heat of vapori-
zation of water is 970 Btu per pound, how

* Some of the energy added to the material leaves
the material as external work and has no effect on
the internal energy of the material. When the
pressure is constant, the amount of external work
done is proportional to the change in volume.
External work is discussed in detail later.

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE 21

much heat is requited to vaporize 3 gal of water
at the saturation temperature of 212° F?

Solution. Total
weight of water M

Applying Equation
2-10, Q L

3 gal x 8.33 lb/gal
= 25 lb

25 lb x 970 Btu/lb
= 24,250 Btu

Example 2-14. One gallon of water at
200° F in an open container absorbs 1200 Btu.
How much water is vaporized?

Solution. Since the saturation temperature
of water at atmospheric pressure is 212° F, the
entire mass of the water must be raised to this
temperature before any water will vaporize

Weight of 1 gal of water

Applying Equation 2-8,
the heat required to raise
the temperature of the
water from 200° F to
212° F, 0,

Heat available to vapor-
ize some portion of the
water

Rearranging and apply-
ing Equation 2-10, the
weight of water vaporized,
M

= 8.33 lb

8.33 lb x
x
= 100 Btu

12°

1200 - 100
1100 Btu

1100
970

1.135 lb or
0.136 gal

Example 2-15. If 5000 Btu are removed
from 8 lb of saturated steam at atmospheric
pressure, how much of the steam will condense
into water?

Solution. By rearrang-
ing and applying Equation
2-10,

M =

5000 Btu
970 Btu/lb
= 5.15 lb

2-34. Superheat — the Sensible Heat of a
Vapor. Once a liquid has been vaporized,
the temperature of the resulting vapor can be
further increased by the addition of heat. The
heat added to a vapor after vaporization is the
sensible heat of the vapor, more commonly
called superheat. When the temperature of a
vapor has been so increased above the satur-
ation temperature, the vapor is said to be
superheated and is called a superheated vapor.
Superheated vapors are discussed at length in
another chapter.
2-35. Total Heat. The total heat of a

material at any particular condition is the sum
total of all the sensible and latent heat required
to bring it to that condition from an initial
condition of Absolute Zero.*

Example 2-16. Compute the total heat
content of 1 lb of steam at 212° F.

Solution. The total heat of 1 lb of saturated
steam is the sum of the following heat quantities :
(a) To raise the temperature of 1 lb of ice from
-460° F to 32° F,

= 1 x 0.5 x

[32 - (-460)]
= 1 x 0.5 x 492
= 246 Btu

(6) To melt 1 lb of ice at 32° F into water at
32° F,

applying Equation 2-9, Q L = 1 x 144

= 144 Btu

applying Equation
2-8,

= 1 x 1 x

(212 - 32)
= 1 x 1 x 180
= 180 Btu

Q L = 1 x 970
= 970 Btu

(d) To vaporize 1 lb of water,
applying Equation 2-10,

(e) Summation:
Sensible heat of the solid = 246 Btu
Latent heat of fusion = 144 Btu
Sensible heat of the liquid = 180 Btu
Latent heat of vaporization = 970 Btu

Total heat of 1 lb

of steam = 1540 Btu

Through the use of a temperature-heat
diagram, the solution to Example 2-15 is shown
graphically in Fig. 2-4.

2-36. Mechanical Energy Equivalent.

Normally the external energy of a body is ex-
pressed in mechanical energy units (work),
whereas the internal energy of a body is ex-
pressed in heat energy units.

The fact that internal energy is usually ex-
pressed in heat energy units gives rise to the
definition of heat as molecular or internal energy.
As previously stated, from a thermodynamic

* The total heat of a material is commonly known
as "enthalpy," and is computed from some
arbitrarily selected zero point rather than from
Absolute Zero. See Section 4-18.

22 PRINCIPLES OF REFRIGERATION

300

212
200

100

-400
-460

- .8

Ss" m
■= 8.

J'S E

#I~ I .3- 3i£Z

M *

J 1

Constant temperature

C/5

r 7

Constant temperature

J L

100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700
Heat content (Btu)

Fig. 2-4. Graphical analysis of the relationship of heat content to the temperature and state of a
material.

point of view energy is heat energy only when
it is in transition from one body to another
because of a difference in temperature between
the two bodies. Once the energy flows into a
body it becomes "stored" thermal energy.
Hence, thermodynamically speaking, internal
energy is not heat but thermal energy in storage.

Not all the heat energy flowing into a body is
stored in the body as internal energy. In many
instances, some or all of the energy flowing into
the body passes through or leaves the body as
work (mechanical energy). This is made clear
in another section.

Furthermore, up to this point it has been
assumed that the internal energy of a body is
increased only by the addition of heat energy
directly, as from a flame or some other heat
source. However, this is not the case. The
internal or molecular energy of a body may also
be increased when work is done on the body.
That is, the mechanical energy of the work done
on a body may be converted to the internal
energy of the body. For example, the head of a
nail struck by a hammer will become warm as a

part of the mechanical energy of the hammer
blow is converted to the internal kinetic energy
of the nail head. As the molecules of the metal
that make up the nail head are jarred and agi-
tated by the blow of the hammer, their motion
or velocity is increased and the temperature of
the nail head increases. If a wire is bent rapidly
back and forth, the bent portion of the wire
becomes hot because of the agitation of the
molecules. Also, everyone is familiar with the
increase in temperature which is brought about
by the friction of two surfaces rubbing together.

Often the external energy of a body is con-
verted to internal energy and vice versa. For
example, a bullet speeding toward a target has
kinetic energy because of its mass and velocity.
At the time of impact with the target, the bullet
loses its velocity and a part of its kinetic energy
is imparted to the molecules of both the bullet
and the target so that the internal energy of
each is increased.

Since heat energy is often converted into
mechanical energy (work) and vice versa, and
since it is often desirable to express both the

MATTER, INTERNAL ENERGY, HEAT, TEMPERATURE 23

internal and external energies of a body in
terms of the same energy unit, a factor which
can be used to convert from one energy unit
to the other is useful.

It has been determined by experiment that
one Btu of heat energy is equivalent to 778
ft-lb of mechanical energy, that is, one Btu is the
amount of heat energy required to do 778 ft-lb
of work. This quantity is known as the mecha-
nical energy equivalent and is usually repre-
sented in equations by the symbol J.

To convert energy in Btu into energy in

foot-pounds, the energy in Btu is multiplied by

778 and, to convert energy in foot-pounds to

energy in Btu, the energy in foot-pounds is

divided by 778. Expressed as equations, these

relationships become

W
Q - -7 (2-11)

and

W = Q xJ

(2-12)

where Q = the quantity of heat energy in Btu
W = mechanical energy or work in foot-
pounds
J = the mechanical energy equivalent
of heat

Example 2-17. Convert 36,000 ft-lb of
mechanical energy into heat energy units.

36,000
Solution. Applying Equa- Q =

tion2-ll,

778
= 46.3 Btu

Example 2-18. Express 12 Btu of heat
energy as work in mechanical energy units.

Solution. Applying
Equation 2-12,

W= 12 x 778
= 9336 ft-lb

PROBLEMS

1. A Fahrenheit thermometer reads 85°. What
is the temperature in degrees Centigrade?

Ans. 29.44° C

2. Convert 90° Centigrade to degrees Fahren-
heit. Ans. 194° F

3. The temperature of a gas is 40° F. What is
its temperature on the Rankine scale?

Ans. 500° R

4. The temperature of the suction vapor enter-
ing a refrigeration compressor is —20° F. What
is the temperature of the vapor in degrees
Rankine? Ans. 440° R

5. Thirty gallons of water are heated from 75° F
to 180° F. Determine the quantity of heat
required? Ans. 26,240 Btu

6. In a certain industrial process, 5000 gal of
water are cooled from 90° F to 55° F each hour.
Determine the quantity of heat which must be
removed each hour to produce the required
cooling. Ans. 1,457,750 Btu

7. Calculate the quantity of heat which must
be removed from 60 gal of water in order to
cool the water from 42° F and freeze it into
ice at 32° F. Ans. 77,000 Btu

8. If 12,120 Btu are added to 3 gal of water at
200° F, what fraction of the water in pounds
will be vaporized? Ans. 9.4 lb or 1.13 gal

9. Twenty-five pounds of ice are placed in
10 gal of water and allowed to melt. Assuming
that there is no loss of heat to the surroundings,
if the initial temperature of the water is 80° F,
to what temperature will the water be cooled
by the melting of the ice ? Ans. 35.7°

10. A gas expanding in a cylinder does 25,000
ft-lb of work on the piston. Determine the
quantity of heat required to do the work.

Ans. 32.13 Btu

Thermodynamic
Processes

3-1. The Effects of Heat on Volume. When
either the velocity of the molecules or the degree
of molecular separation is increased by the
addition of heat, the mean distance between the
molecules is increased and the material expands
so that a unit weight of the material occupies a
greater volume. This effect is in strict accord-
ance with the theory of increased or decreased
molecular activity as described earlier. Hence,
when heat is added to or removed from an
unconfined material in any of the three physical
states, it will expand or contract, respectively.
That is, its volume will increase or decrease with
the addition or removal of heat.

One of the few exceptions to this rule is water.
If water is cooled, its volume will decrease
normally until the temperature of the water
drops to 39.2° F. At this point, water attains
its maximum density and, if further cooled, its
volume will again increase. Furthermore, after
being cooled to 32° F, it will solidify and the
solidification will be accompanied by still
further expansion. In fact, 1 cu ft of water will
freeze into approximately 1.085 cu ft of ice.
This accounts for the tremendous expansive
force created during solidification which is
sufficient to burst steel pipes or other restraining
vessels.

The peculiar behavior of water as it solidifies
appears to contradict the general laws govern-
ing molecular activity as described previously.
However, this is not the case. The unusual
behavior of water is explained by the hypothesis

that, although the molecules of water are
actually closer together in the solid state than
they are in the liquid state, they are grouped
together to form crystals. It is the relatively
large spaces between the crystals of the solid,
rather than any increase in the mean distance
between the molecules, which accounts for the
unusual increase in volume during solidification.
This is true also for crystalline solids other than
ice.

3-2. Expansion of Solids and Liquids. When
a solid or a liquid is heated so that its tempera-
ture is increased, it will expand a given amount
for each degree of temperature rise. As stated
earlier, many temperature measuring devices
are based upon this principle. The amount of
expansion which a material experiences with
each degree of temperature rise is known as its
coefficient of expansion. The coefficient of
expansion is different for every material, and
moreover it will vary for any particular material
depending upon the temperature range in
which the change occurs.

Since solids and liquids are not readily com-
pressible, if a solid or a liquid is restrained or
confined so that its volume is not allowed to
change normally with a change in temperature,
tremendous pressures are created within the
material itself and upon the restraining bodies,
which is likely to cause buckling or rupturing
of either the material, the restraining bodies, or
both. To provide for the normal expansion and
contraction occurring with temperature changes,
expansion joints are built into highways,
bridges, pipelines, etc. Likewise, liquid con-
tainers are never completely filled. Space must
be allowed for the normal expansion. Other-
wise the tremendous expansive forces generated
by a temperature increase will cause the con-
taining vessel to rupture, sometimes with
explosive force.

3-3. Specific Volume. The specific volume
of a material is the volume occupied by a 1 lb
mass of the material. Each material has a
different specific volume and, because of the
change in volume which accompanies a change
in temperature, the specific volume of every
material varies somewhat with the temperature
range. For instance, at 40° F, 1 lb of water has
a specific volume of 0.01602 cu ft, whereas the
volume occupied by 1 lb of water at 80° F is
0.01608 cu ft.

THERMODYNAMIC PROCESSES 25

3-4. Density. The density of a material is the
weight in pounds of 1 cu ft of the material.
Density is die reciprocal of specific volume, that
is, the specific volume divided into one. The
density of any material, like specific volume,
varies with the temperature, but in the opposite
direction. For example, at 40° F, the density
of water is 62.434 lb per cubic foot (1/0.01602),
whereas water at 80° F has a density of 62.20
lb per cubic foot (1/0.01608). Since density and
specific volume are reciprocals of each other,
as one increases the other decreases. The
density and/or the specific volume of many
common materials can be found in various
tables.

The relationship between density and specific
volume is given by the following equations:

p= ~v

(3-1)

V = -

(3-2)

V = M x v

(3-3)

M= V x P

(3-4)

where v = the specific volume in cubic feet per
pound (cu ft/lb)
p = the density in pounds per cubic foot

(lb/cu ft)
V = the total volume in cubic feet
M = the total weight in pounds

Example 3-1 . If the specific volume of dry
saturated steam at 212° F is 26.80 cuft per
pound, what is the density of the steam?

Solution. Applying
Equation 3-1, the density p

26.80
= 0.0373 lb/cu ft

Example 3-2. The basin of a cooling tower,
measuring 5 ft x 4 ft x 1 ft, is filled with water.
If the density of the water is 26.8 lb per cubic
foot, what is the total weight of the water in the
basin?

Solution. The total volume V 5 ft x 4 ft

x 1ft
= 20 cu ft
20 X 26.8

- 536 lb

Applying Equation 3-4, the
total weight of water M

3-5. Pressure-Temperature-Volume Rela-
tionships of Gases. Because of its loose
molecular structure, the change in the volume of
a gas as the gas is heated or cooled is much
greater than that which occurs in the case of a
solid or a liquid. In the following sections, it
will be shown that a gas may change its con-
dition in a number of different ways and that
certain laws have been formulated which govern
the relationship between the pressure, temper-
ature, and volume of the gas during these
changes. It should be noted at the outset that
in applying the fundamental gas laws it is
always necessary to use absolute pressures and
absolute temperatures in degrees Rankine.
Further, in studying the following sections, it
should be remembered that a gas always
completely fills any container.

The relationship between the pressure,
temperature, and volume of a gas is more
easily understood when considered through a
series of processes in which the gas passes
from some initial condition to some final con-
dition in such a way that only two of these
properties vary during any one process, whereas
the third property remains unchanged or con-
stant.

3-6. Temperature-Volume Relationship at
a Constant Pressure. If a gas is heated
under such conditions that its pressure is kept
constant, its volume will increase 1/492 of its
volume at 32° F for each 1° F increase in its
temperature. Likewise, if a gas is cooled at a
constant pressure, its volume will decrease 1/492
of its volume at 32° F for each 1 ° F decrease in
its temperature.

In order to better visualize a constant pressure
change in condition, assume that a gas is con-
fined in a cylinder equipped with a perfectly
fitting, frictionless piston (Fig. 3-la). The
pressure of the gas is that which is exerted on the
gas by the weight of the piston and by the weight
of the atmosphere on top of the piston. Since
the piston is free to move up or down in the
cylinder, the gas is allowed to expand or con-
tract, that is, to change its volume in such a way
that the pressure of the gas remains constant.
As the gas is heated, its temperature and volume
increase and the piston moves- upward in the
cylinder. As the gas is cooled, its temperature
and volume decrease and the piston moves down-
ward in the cylinder. In either case, the pressure

26 PRINCIPLES OF REFRIGERATION

Perfectly fitting

frictionless

piston

: Piston

■■?< P = 100 psia
£fr = 500*R J&
&V=Icuft

[tttl

Hi i Pistoh||

ft? '.■ 'Of/^'"""-"-"^

i£f P = 100 psia $

$v;T=1000*R 4.

Volu
chai

Volume
change

fTTT

Heat
added

Heat
removed

(a) (b) (c)

Fig. 3-1. Constant pressure process, (a) Gas confined In a cylinder with a perfectly fitting, frictionless piston,
(b) As gas is heated, both the temperature and the volume of the gas Increase. The increase in volume is
exactly proportional to the increase in absolute temperature, (c) As gas is cooled, both the temperature and
the volume of the gas decrease. The decrease in volume is exactly proportional to the decrease in absolute
temperature.

of the gas remains the same or unchanged during
the heating or cooling processes.
3-7. Charles' Law for a Constant Pressure
Process. Charles' law for a constant pressure
process states in effect that, when the pressure
of the gas remains constant, the volume of the
gas varies directly with its absolute temperature.
Thus, if the absolute temperature of a gas is
doubled while its pressure is kept constant, its
volume will also be doubled. Likewise, if the
absolute temperature of a gas is reduced by
one-half while the pressure is kept constant, its
volume will also be reduced by one-half. This
relationship is illustrated in Figs. 3-16 and
3-lc.

Charles' law for a constant pressure process
written as an equation is as follows: When the
pressure is kept constant,

TrV t = T^

(3-5)

where T ± = the initial temperature of the gas in

degrees Rankine
r 2 = the final temperature of the gas in

degrees Rankine
Vx = the initial volume of the gas in

cubic feet
V s = the final volume of the gas in cubic

feet

When any three of the preceding values are
known, the fourth may be calculated by applying
Equation 3-5.

Example 3-3. A gas, whose initial tem-
perature is 520° R and whose initial volume is
5 cu ft, is allowed to expand at a constant
pressure until its volume is 10 cu ft. Determine
the final temperature of the gas in degrees
Rankine.

520 x 10

Solution. By rearranging and
applying Equation 3-5, the final
temperature of the gas T 2

= 1040° R

Example 3-4. A gas, having an initial
temperature of 80° F, is cooled at a constant
pressure until its temperature is 40° F. If the
initial volume of the gas is 8 cu ft, what is its
final volume?

Solution. Since the temperatures are given
in degrees Fahrenheit, they must be converted
to degrees Rankine before being substituted in
Equation 3-5.

By rearranging and apply- _ ^i^i
ing Equation 3-5, the final T t

volume V 2 = 500 x 8

540
= 7.4074 cu ft

THERMODYNAMIC PROCESSES 27

3-8. Pressure-Volume Relationship at a
Constant Temperature. When the volume
of a gas is increased or decreased under such
conditions that the temperature of the gas does
not change, the absolute pressure will vary
inversely with the volume. Thus, when a gas is
compressed (volume decreased) while its tem-
perature remains unchanged, its absolute
pressure will increase in proportion to the
decrease in volume. Similarly, when a gas is
expanded at a constant temperature, its abso-
lute pressure will decrease in proportion to the
increase in volume. This is a statement of
Boyle's law for a constant temperature process
and is illustrated in Figs. 3-2a, 3-26, and 3-2c.
It has been previously stated that the mole-
cules of a gas are flying about at random and at
high velocities and that the molecules of the
gas frequently collide with one another and with
the walls of the container. The pressure exerted
by the gas is a manifestation of these molecular
collisions. Billions and billions of gas mole-
cules, traveling at high velocities, strike the
walls of the container during each fraction of a
second. It is this incessant molecular bombard-
ment which produces the pressure that a gas
exerts upon the walls of its container. The
magnitude of the pressure exerted depends upon
the force and frequency of the molecular im-
pacts upon a given area. The greater the force

and frequency of the impacts, the greater is the
pressure. The number of molecules confined in
a given space and their velocity will, of course,
determine the force and the frequency of the
impacts. That is, the greater the number of
molecules (the greater the quantity of gas) and
the higher the velocity of the molecules (the
higher the temperature of the gas), the greater
is the pressure. The force with which the mole-
cules strike the container walls depends only
upon the velocity of the molecules. The higher
the velocity the greater is the force of impact.
The greater the number of molecules in a given
space and the higher the velocity the more often
the molecules will strike the walls.

When a gas is compressed at a constant
temperature, the velocity of the molecules re-
mains unchanged. The increase in pressure
which occurs is accounted for by the fact that the
volume of the gas is diminished and a given
number of gas molecules are confined in a smaller
space so that the frequency of impact is greater.
The reverse of this holds true, of course, when
the gas is expanded at a constant temperature.

Any thermodynamic process which occurs in
such a way that the temperature of the working
substance does not change during the process is
called an isothermal (constant temperature)
process.

Boyle's law for a constant temperature process

Fig. 3-2. Constant tempera-
ture process, (a) Initial condi-
tion, (b) Constant temperature
expansion — volume change is
inversely proportional to the
change in absolute pressure.
Heat must be added during
expansion to keep tempera-
ture constant, (c) Constant
temperature compression —
volume change is inversely
proportional to the change in
absolute pressure. Heat must
be removed during com-
pression to keep temperature
constant.

mi: piston: mi

-■,■ 1 I-'
1 ■ 1 '

'■:■■: II ■■;'.:.• '■
"■^:'l l:---'v
-__-l-L_

i
i

i
l
i

miipiswiiu

<p=ibopsi |

V=lcuft
! T=70*F

P * 50 psi
. V«2cuft $
? T=70*F ;

HI|liPiston:i||

iP m 200 psi<2
IV = 0.5cuft1
!§J>.70*F

Heat
(1

'If

idded

H'l

Heat re
U

moved
)

28 PRINCIPLES OF REFRIGERATION

Volume = 1 cu ft

Pressure = 100 psia

Temperatures 500° R,

Volume = 1 cu ft @>

Pressure = 200 psia

Temperature= 1000° R

Volume = 1 cu ft

Pressure = 50 psia

Temperature = 250' R

TTT

Heat added

Heat removed

(a) (b) (c)

Fig. 3-3. Constant volume process, (a) Initial condition, (b) The absolute pressure increases in direct
proportion to the increase in absolute temperature, (c) The absolute pressure decreases in direct proportion
to the decrease in absolute temperature.

is represented by the following equation: if
the temperature is constant,

PiVi = P*r* (3-6)

where P x = the initial absolute pressure
P a = the final absolute pressure
V x = the initial volume in cubic feet
K 2 = the final volume in cubic feet

Example 3-5. Five pounds of air are ex-
panded at a constant temperature from an
initial volume of 5 cu ft to a final volume of
10 cu ft. If the initial pressure of the air is
20 psia, what is the final pressure in psia?

20 x 5

initial pressure of the gas is 3000 psfa, determine
the final pressure in psig.

Solution. By rearranging and
applying Equation 3-6, the final
pressure P 2

10
= 10 psia

Example 3-6. Four cubic feet of gas are
allowed to expand at a constant temperature
from an initial pressure of 1500 psfa to a final
pressure of 900 psfa. Determine the final
volume of the gas.

_ PiVi

P*
_ 1500 x 4

900
= 6.67 cu ft

Solution. By rearranging and
applying Equation 3-6, the
final volume V z

Example 3-7. A given weight of gas, whose
initial volume is 10 cu ft, is compressed iso-
thermally until its volume is 4 cu ft. If the

Solution. By rearranging
and applying Equation 3-6,
the final pressure P t

PiVi

3000 x 10
4
= 7500 psfa

Dividing by 144

7500
144

Subtracting the atmos-
pheric pressure

= 52.08 psia
= 52.08 - 14.7
= 37.38 psig

3-9. Pressure-Temperature Relationship
at a Constant Volume. Assume that a gas
is confined in a closed cylinder so that its
volume cannot change as it is heated or cooled
(Fig. 3-3a). When the temperature of a gas is
increased by the addition of heat, the absolute
pressure will increase in direct proportion to the
increase in absolute temperature (Fig. 3-3b).
If the gas is cooled, the absolute pressure of the
gas will decrease in direct proportion to the
decrease in absolute temperature (Fig. 3-3c).

Whenever the temperature (velocity of the
molecules) of a gas is increased while the volume
of the gas (space in which the molecules are
confined) remains the same, the magnitude of
the pressure (the force and frequency of mole-
cular impacts on the cylinder walls) increases.
Likewise, when a gas is cooled at a constant
volume, the force and frequency of molecular
impingement on the walls of the container

diminish and the pressure of the gas will be less
than before. The reduction in the force and the
frequency of molecular impacts is accounted
for by the reduction in molecular velocity.
3-10. Charles' Law for a Constant Volume
Process. Charles' law states in effect that
when a gas is heated or cooled under such con-
ditions that the volume of the gas remains
unchanged or constant, the absolute pressure
varies directly with the absolute temperature.
Charles' law may be written as the following
equation: when the volume is the same,

THERMODYNAMIC PROCESSES 29

is used, then V will become the specific volume
v, and Equation 3-8 may be written:

Vi = Vi

(3-7)

where T x = the initial temperature in degrees

Rankine
r 2 = the final temperature in degrees

Rankine
P x = the initial pressure in pounds per

square inch absolute
P 2 = the final pressure in pounds per

square inch absolute

Example 3-8. A certain weight of gas con-
fined in a tank has an initial temperature of
80° F and an initial pressure of 30 psig. If the
gas is heated until the final gage pressure is
50 psi, what is the final temperature in degrees
Fahrenheit?

Solution. By rearrang-
ing and applying Equa-
tion 3-7,

Converting Rankine to
Fahrenheit

T, =

T, xi».

(80 + 460)
_ x(50 + 14.7)
(30 + 14.7)
= 782° R

782 - 460
= 322° F

3-11. The General Gas Law. Combining
Charles' and Boyle's laws produces the following
equation:

P -P= P -P 0-8)

Equation 3-8 is a statement that for any given
weight of a gas, the product of the pressure in
psfa and the volume in cubic feet divided by the
absolute temperature In degrees Rankine will
always be a constant. The constant, of course,
will be different for different gases and, for any
one gas, will vary with the weight of gas involved.
However, if, for any one gas, the weight of 1 lb

Pv
T

(3-9)

where R = the gas constant

The gas constant R is different for each gas.
The gas constant for most common gases can
be found in tables. A few of these are given
in Table 3-1.

Multiplying both sides of Equation 3-9 by M
produces

PMv = MRT
but since

Mv = V
then

PV = MRT (3-10)

where P = the pressure in psfa
V = the volume in ft 3
M = the mass in pounds
R = the gas constant
T = the temperature in ° R

Equation 3-10 is known as the general gas law
and is very useful in the solution of many
problems involving gases. Since the value of R
for most gases can be found in tables, if any three
of the four properties, P, V, M, and T, are known,
the fourth property can be determined by Equa-
tion 3-10. Notice that the pressure must be in
pounds per square foot absolute.

Example 3-9. The tank of an air compres-
sor has a volume of 5 cu ft and is filled with air
at a temperature of 100° F. If a gage on the
tank reads 151.1 psi, what is the weight of the
air in the tank?

Solution. From
Table 3-1, R for air

By rearranging
and applying Equa-
tion 3-10, the weight
of air M

= 53.3

(151.1 + 14.7)

x 144x5

~ 53.3 x (100 + 460)
_ 165.8 x 144 x 5

53.3 x 560
= 4 lb

Example 3-10. Two pounds of air have a
volume of 3 cu ft. If the pressure of the air is
135.3 psig, what is the temperature of the air in
degrees Fahrenheit?

30 PRINCIPLES OF REFRIGERATION

Solution. From

Table 3-1, R for air

= 53.3

By rearranging
and applying Equa-
tion 3-10, the tem-
perature of the air

T- PV

~ MR

(135.3 + 14.7)

x 144x3

in °R,

2 x 53.3

150 x 144 x 3

2 x 53.3

= 607.9° R

Converting to
Fahrenheit,

t = 607.9 - 460
= 147.9° F

3-12. External Work. Whenever a material
undergoes a change in volume, work is done. If
the volume of the material increases.work is done
by the material. If the volume of the material
decreases, work is done on the material. For
example, consider a certain weight of gas con-
fined in a cylinder equipped with a movable
piston (Fig. 3-la). As the gas is heated, its
temperature increases and it expands, moving
the piston upward in the cylinder against the
pressure of the atmosphere. Work is done in
that the weight of the piston is moved through a
distance(Fig. 3-1A).* The agency doing the work
is the expanding gas.

In order to do work, energy is required
(Section 1-12). In Fig. 3-lZ>, the energy required
to do the work is supplied to the gas as the gas
is heated by an external source. It is possible,
however, for a gas to do external work without
the addition of energy from an external source.
In such cases, the gas does the work at the
expense of its own energy. That is, as the gas
expands and does work, its internal kinetic energy
(temperature) decreases in an amount equal to
the amount of energy required to do the work.

When a gas is compressed (its volume de-
creased), a certain amount of work must be done
on the gas in order to compress it. And, an
amount of energy equal to the amount of work
done will be imparted to the molecules of the
gas during the compression. That is, the
mechanical energy of the piston motion will be
transformed into the internal kinetic energy of
the gas (molecular motion) and, unless the gas is
cooled during the compression, the temperature
of the gas will increase in proportion to the

* Some work is done, also, in overcoming friction
and in overcoming the pressure of the atmosphere.

amount of work done. The increase in the tem-
perature of a gas as the gas is compressed is a
common phenomenon and may be noted by
feeling the valve stem of a tire being filled with a
hand pump, or the head of an air compressor,
etc.

3-13. The General Energy Equation. The
law of conservation of energy clearly indicates
that the energy transferred to a body must be
accounted for in its entirety. It has been shown
that some part (or all) of the energy taken in by a
material may leave the material as work, and
that only that portion of the transferred energy
which is not utilized to do external work remains
in the body as "stored thermal energy." It is
evident then that all of the energy transferred to
a body must be accounted for in some one or in
some combination of the following three ways:

(1) as an increase in the internal kinetic energy,

(2) as an increase in the internal potential energy,
and (3) as external work done. The general
energy equation is a mathematical statement of
this concept and may be written:

AQ = AK + AP + AW

(3-11)

where AQ = the heat energy transferred to the
material in Btu
AK = that fraction of the transferred
energy which increases the in-
ternal kinetic energy
AP = that fraction of the transferred
energy which increases the in-
ternal potential energy
A W = that fraction of the transferred
energy which is utilized to do
external work

The Greek letter, A (delta), used in front of a
term in an equation identifies a change of
condition. For instance, where K represents
the internal kinetic energy, AK represents the
change in the internal kinetic energy.

Depending upon the particular process or
change in condition that the material undergoes,
any of the terms in Equation 3-11 may have any
value either positive or negative, or any may be
equal to zero. This will be made clear later.
3-14. External Work of a Solid or Liquid.
When heat added to a material in either the
solid or liquid state increases the temperature
of the material, the material expands somewhat
and a small amount of work is done. However,

THERMODYNAMIC PROCESSES 31

the increase in volume and the external work
done is so slight that the portion of the trans-
ferred energy which is utilized to do external
work or to increase the internal potential energy
is negligible. For all practical purposes, it can
be assumed that all the energy added to a solid
or a liquid during a temperature change in-
creases the internal kinetic energy. None leaves
the material as work and none is set up as an
increase in the internal potential energy. In
this instance, both AP and AW of Equation
3-11 are equal to zero and, therefore, AQ is
equal to AK .

When a solid melts into the liquid phase,
the change in volume is again so slight that the
external work done may be neglected. Further-
more, since the temperature also remains
constant during the phase change, none of the
transferred energy increases the internal kinetic
energy. All the energy taken in by the melting
solid is set up as an increase in the internal
potential energy. Therefore, AK and AW are
both equal to zero and A Q is equal to AP.

This is not true, however, when a liquid
changes into the vapor phase. The change in
volume that occurs and therefore the external
work done as the liquid changes into a vapor is
considerable. For example, when 1 lb of water
at atmospheric pressure changes into a vapor,
its volume increases from 0.01671 cu ft to 26.79
cu ft. Of the 970.4 Btu required to vaporize 1
lb of water, approximately 72 Btu of this energy
are required to do the work of expanding
against the pressure of the atmosphere. The
remainder of the energy is set up in the vapor as
an increase in the internal potential energy. In
this instance, only AK is equal to zero, so that
AQ is equal to AP plus AW.
3-15. "Ideal" or "Perfect" Gas. The various
laws governing the pressure-volume-temperature
relationships of gases as discussed in this chapter
apply with absolute accuracy only to a hypo-
thetical "ideal" or "perfect" gas. A "perfect"
gas is described as one in such a condition that
there is no interaction between the molecules of
the gas. The molecules of such a gas are
entirely free and independent of each other's
attractive forces. Hence, none of the energy
transferred either to or from an ideal gas has
any effect on the internal potential energy.

The concept of an ideal or perfect gas greatly
simplifies the solution of problems concerning

the changes in the condition of a gas. Many
complex problems in mechanics are made simple
by die assumption that no friction exists, the
effects of friction being considered separately.
The function of an ideal gas is the same as that
of the frictionless surface. An ideal gas is
assumed to undergo a change of condition
without internal friction, that is, without the
performance of internal work in the overcoming
of internal molecular forces.

The idea of internal friction is not difficult to
comprehend. Consider that a liquid such as oil
will not flow readily at low temperatures. This
is because of the internal friction resulting from
strong intermolecular forces within the liquid.
However, as the liquid is heated and the mole-
cules gain additional energy, the intermolecular
forces are overcome somewhat, internal friction
diminishes, and the liquid flows more easily.

Vaporization of the liquid, of course, causes
a greater separation of the molecules and
brings about a substantial reduction in internal
friction, but some interaction between the
molecules of the vapor still exists. In the
gaseous state, intermolecular forces are greatest
when the gas is near the liquid phase and
diminish rapidly as the gas is heated and its
temperature rises farther and farther above the
saturation temperature. A gas approaches the
ideal state when it reaches a condition such that
the interaction between the molecules and
hence, internal friction, is negligible.

Although no such thing as an ideal or perfect
gas actually exists, many gases, such as air,
nitrogen, hydrogen, helium, etc., so closely
approach the ideal condition that any errors
which may result from considering them to be
ideal are of no consequence for all practical
purposes.

Although it is important that the student of
refrigeration understand and be able to apply
the laws of perfect gases, it should be under-
stood that gases as they normally occur in the
mechanical refrigeration cycle are close to the
saturation curve, that is, they are vapors, and
do not even approximately approach the
condition of an ideal or perfect gas.* They
follow the gas laws in only a very general way,

* A vapor is sometimes defined as a gas at a
condition close enough to the saturation curve so
that it does not follow the ideal gas laws even
approximately.

32 PRINCIPLES OF REFRIGERATION

and therefore the use of the gas laws to deter-
mine the pressure-volume-temperature relation-
ships of such vapors will result in considerable
inaccuracy. In working with vapors, it is usually
necessary to use values which have been deter-
mined experimentally and are tabulated in
saturated and superheated vapor tables. These
tables are included as a part of this textbook
and are discussed later.

3-16. Processes for Ideal Gases. A gas is
said to undergo a process when it passes from
some initial state or condition to some final
state or condition. A change in the condition of
a gas may occur in an infinite number of ways,
only five of which are of interest. These are the
(1) constant pressure (isobaric), (2) constant
volume (isometric), (3) constant temperature
(isothermal), (4) adiabatic, and (5) polytropic
processes.

In describing an ideal gas, it has been said
that the molecules of such a gas are so far apart
that they have no attraction for one another,
and that none of the energy absorbed by an
ideal gas has any effect on the internal potential
energy. It is evident, then, that heat absorbed
by an ideal gas will either increase the internal
kinetic energy (temperature) of the gas or it will
leave the gas as external work, or both. Since
the change in the internal potential energy, AP,
will always be zero, the general energy equation
for an ideal gas may be written:

AQ = AK + AW

(3-12)

In order to better understand the energy
changes which occur during the various pro-
cesses, it should be kept in mind that a change in
the temperature of the gas indicates a change in
the internal kinetic energy of the gas, whereas a
change in the volume of the gas indicates work
done either by or on the gas.
3-17. Constant Volume Process. When a
gas is heated while it is so confined that its
volume cannot change, its pressure and tempera-
ture will vary according to Charles' law (Fig.
3-3). Since the volume of the gas does not
change, no external work is done and AW is
equal to zero. Therefore, for a constant
volume process, indicated by the subscript v,

*Qv - Mv

(3-13)

transferred to the gas increases the internal
kinetic energy of the gas. None of the energy
leaves the gas as work.

When a gas is cooled (heat removed) while
its volume remains constant, all the energy
removed is effective in reducing the internal
kinetic energy of the gas. It should be noted
that in Equation 3-12, AQ represents heat
transferred to the gas, AK represents an in-
crease in the internal kinetic energy, and AW
represents work done by the gas. Therefore,
if heat is given up by the gas, AQ is negative.
Likewise, if the internal kinetic energy of the gas
decreases, AK is negative and, if work is done on
the gas, rather than by it, A W is negative. Hence,
in Equation 3-13, when the gas is cooled, both
AQ and AK are negative.
3-18. Constant Pressure Process. If the
temperature of a gas is increased by the addition
of heat while the gas is allowed to expand so
that its pressure is kept constant, the volume of
the gas will increase in accordance with Charles'
law (Fig. 3-1). Since the volume of the gas
increases during the process, work is done by
the gas at the same time that its internal energy
is increased. Hence, while one fraction of the
transferred energy increases the store of internal
kinetic energy, another fraction of the trans-
ferred energy leaves the gas as work. For a
constant pressure process, identified by the
subscript/!, the energy equation may be written

AQ P =AK 1I +AW

(3-14)

Equation 3-13 is a statement that during a
constant volume process all of the energy

3-19. Specific Heat of Gases. The quantity
of heat required to raise the temperature of
1 lb of a gas 1° F while the volume of the gas
remains constant is known as the specific heat
at a constant volume (C„). Similarly, the
quantity of heat required to raise the tempera-
ture of 1 lb of a gas 1° F while the gas expands
at a constant pressure is called the specific
heat at a constant pressure (C„). For any
particular gas, the specific heat at a constant
pressure is always greater than the specific heat
at a constant volume. The reason for this is
easily explained.

The quantity of energy required to increase
the internal kinetic energy of a gas to the extent
that the temperature of the gas is increased 1 ° F
is exactly the same for all processes. Since,
during a constant volume process, no work is
done, the only energy required is that which

THERMODYNAMIC PROCESSES 33

increases the internal kinetic energy. However,
during a constant pressure process, the gas
expands a fixed amount for each degree of
temperature rise and a certain amount of
external work is done. Therefore, during a
constant pressure process, energy to do the
work that is done must be supplied in addition
to that which increases the internal kinetic
energy. For example, the specific heat of air
at a constant volume is 0.169 Btu per pound,
whereas the specific heat of air at a constant
pressure is 0.2375 Btu per pound. For either
process, the increase in the internal energy of
the air per degree of temperature rise is 0.169
Btu per pound. For the constant pressure pro-
cess, the additional 0.068S Btu per pound
(0.2375 — 0.169) is the energy required to do
the work resulting from the volume increase
accompanying the temperature rise.

The specific heat of a gas may take any value
either positive or negative, depending upon the
amount of work that the gas does as it expands.
3-20. The Change in Internal Kinetic
Energy. During any process in which the
temperature of the gas changes, there will be a
change in the internal kinetic energy of the gas.
Regardless of the process, when the tempera-
ture of a given weight of gas is increased or
decreased, the change in the internal kinetic
energy can be determined by the equation

AK = MC v (t t - fj

(3-15)

where AK = the increase in the internal kinetic
energy in Btu
M = the weight in pounds
C v = constant volume specific heat
t t = the final temperature
f x = the initial temperature

Note. The temperature may be in either
Fahrenheit or Rankine, since the difference in
temperature will be the same in either case as
long as the units are consistent.

Example 3-11. The temperature of 5 lb
of air is increased by the addition of heat from
an initial temperature of 75° F to a final tem-
perature of 140° F. If C„ for air is 0.169 Btu,
what is the increase in the internal energy?

Solution. Applying =5 x 0.169
Equation 3-15, the in- x (140 - 75)

crease in internal kinetic = 5 x 0.169 x 65

energy AK = 54.9 Btu

Example 3-12. Twelve pounds of air are
cooled from an initial temperature of 95° F to
a final temperature of 72° F. Compute the
increase in the internal kinetic energy.

Solution. Apply-
ing Equation 3-15,
the increase in in-
ternal kinetic energy
AK

= 12 x 0.169

x (72 - 95)
= 12 x 0.169 x (-23)
= -46.64 Btu

In Example 3-12, AK is negative, indicating
that the gas is cooled and that the internal
kinetic energy is decreased rather than increased.
3-21. Heat Transferred during a Constant
Volume Process. For a constant volume
process, since

then

AQ V = AK V
AQ V = MC v {ti - /,) (3-16)

Example 3-13. If, in Example 3-11, the gas
is heated while its volume is kept constant,
what is the quantity of heat transferred to the
gas during the process ?

Solution. Apply-
ing Equation 3-16,
the heat transferred
to the gas,

Alternate Solution.
From Example
3-12,

Since
AQ V = AK V ,

AQ V = 5 x 0.169

x (140 - 75)
= 5 x 0.169 x 65
= 54.9 Btu

AK V = 54.9 Btu
AQ V = 54.9 Btu

Example 3-14. If, in Example 3-12, the
air is cooled while its volume remains constant,
what is the quantity of heat transferred to the
air during the process?

Solution. From
Example 3-12,
Since AQ„ = AK V .

AK V = -46.67 Btu
AQ V 46.67 Btu

In Example 3-14, notice that since AK V is
negative, indicating a decrease in the internal
kinetic energy, AQ V must of necessity also be
negative, indicating that heat is transferred
from the gas rather than to it.
3-22, External Work during a Constant
Pressure Process. It will now be shown that
the work done during a constant pressure
process may be evaluated by the equation :

W'PiVt-VJ

(3-17)

34 PRINCIPLES OF REFRIGERATION

where W = the work done in foot-pounds
P = the pressure in psfa
Kg = the final volume in cubic feet
V x = the initial volume in cubic feet

Assume that the piston in Fig. 3-lc has an
area of A square feet and that the pressure of
the gas in the cylinder is P pounds per square
foot. Then, the total force exerted on the top of
the piston will be PA pounds, or

F = P xA

Assume now that the gas in the cylinder, having
an initial volume V ly is heated and allowed to
expand to volume V 2 while its pressure is kept
constant. In doing so, the force PA acts through
the distance 1 and work is done. Hence,

but since

then

W = P xA x 1

A x 1 = (V s - VJ
W = P(V 2 -VJ

Example 3-15. One pound of air having
an initial volume of 13.34 cu ft and an initial
temperature of 70° F is heated and allowed to
expand at a constant pressure of 21 17 psfa to a
final volume of 15 cu ft. Determine the amount
of external work in foot-pounds.

Solution. Applying
Equation 3-17, the work in
foot-pounds W

= 2117 x

(15 - 13.34)
= 2117 x 1.66
= 3514 ft-lb

In Equation 3-12, AWS& always given in heat
energy units. By application of the mechanical
energy equivalent (Section 3-16), W in foot-
pounds may be expressed as A W in Btu. The
relationship is

W
AW = — (3-18)

W = AW xJ

(3-19)

Example 3-16. Express the work done in
Example 3-15 in terms of heat energy units.

Solution. Applying Equation _ 3514
3-18, the work in Btu W 778

= 4.52 Btu

3-23. Heat Transferred during a Constant
Pressure Process. According to Equation
3-14, AQ„ the total heat transferred to a gas
during a constant pressure process is equal to

the sum of AK P , the increase in internal kinetic
energy, and AW 9 , the heat energy equivalent of
the work of expansion.

Example 3-17. Compute the total heat
energy transferred to the air during the constant
pressure process described in Example 3-15.

Solution. Con-

verting 70° F to
degrees Rankine,

° R = 70 + 460
= 530° R

Applying
Charles' law,

T — *1 X ^1

2 " v x

Equation 3-5, to
determine the final

_ 530 x 15
13.34

temperature,

= 596" R

Applying Equa-
tion 3-15, the in-
crease in internal

AA: = 1 x 0.169
x (596 -
= 1 x 0.169

530)
x 66

kinetic energy,

= 11.154 Btu

From Example
3-15 and 3-16,

DW, =4.52 Btu

Applying Equa- AQ V = 11.15 + 4.52
tion 3-14, = 15.67 Btu

Since the specific heat at a constant pressure
C P takes into account not only the increase in
internal energy per pound but also the work
done per pound per degree of temperature rise
during a constant pressure expansion, for the
constant pressure process only, AQ P may be
determined by the following equation :

AQ P - MC p (t 2 - tj)

(3-20)

Hence, an alternate solution to Example 3-17 is

Applying Equation AQ V = 1 x 0.2375

x (596 - 530)
= 1 x 0.2375

x 66
= 15.67 Btu

3-20,

3-24. Pressure-Volume (PV) Diagram.
Equation 3-8 is a statement that the thermo-
dynamic state of a gas is adequately described
by any two properties of the gas. Hence, using
any two properties of the gas as mathematical
coordinates, the thermodynamic state of a gas
at any given instant may be shown as a point
on a chart. Furthermore, when the conditions
under which a gas passes from some initial state
to some final state are known, the path that the
process follows may be made to appear as a line
on the chart.

THERMODYNAMIC PROCESSES 35

The graphical representation of a process or
cycle is called a process diagram or a cycle
diagram, respectively, and is a very useful tool
in the analysis and solution of cyclic problems.

Since work is a function of pressure and
volume, when it is the work of a process or
cycle which is of interest, the properties used as
coordinates are usually the pressure and the
volume. When the pressure and volume are
used as coordinates to diagram a process or
cycle, it is called a pressure-volume (PV)
diagram.

To illustrate the use of the PV diagram, a
pressure-volume diagram of the process de-
scribed in Example 3- IS is shown in Fig. 3-4.
Notice that the pressure in psfa is used as the
vertical coordinate, whereas the volume in cubic
feet is used as the horizontal coordinate.

In Example 3-1 5, the initial condition of the
gas is such that the pressure is 21 17 psfa and the
volume is 13.34 cu ft. To establish the initial
state of the gas on the PV chart, start at the
origin and proceed upward along the vertical
pressure axis to the given pressure, 2117 psfa.
Draw a dotted line parallel to the base line
through this point and across the chart. Next,
from the point of origin proceed to the right
along the horizontal volume axis to the given
volume, 13.34 cu ft. Through this point draw a
vertical dotted line across the chart. The inter-
section of the dotted lines at point 1 establishes
the initial thermodynamic state of the gas.

According to Example 3-1 5, the gas is heated
and allowed to expand at a constant pressure
until its volume is IS cu ft. Since the pressure
remains the same during the process, the state
point representing the final state of the gas must
fall somewhere along the line of constant
pressure already established. The exact point
on the pressure line which represents the final
state 2 is determined by the intersection of the
line drawn through the point on the volume axis
that identifies the final volume.

In passing from the initial state 1 to the final
state 2 the air passes through a number of
intermediate thermodynamic states, all of which
can be represented by points which will fall
along line 1 to 2. Line 1 to 2, then, represents
the path that the process will follow as the ther-
modynamic state of the gas changes from 1 to 2,
and is the PV diagram of the process described.

The area of a rectangle is the product of its

4000

_ Final
condition

13 ! 14

v 2

Volume (cubic feet)
Fig. 3-4. Pressure-volume diagram of constant
pressure process. Crosshatched area between
process diagram and base line represents external
work done during the process.

two dimensions. In Fig. 3-4, the area of the
rectangle, 1-2-Kg-^ (crosshatched), is the
product of its altitude P and its base (K g — Fj).
But according to Equation 3-17, the product
P(V t — VJ is the external work done during a
constant pressure process. It is evident then
that the area between the process diagram and
the volume axis is a measure of the external
work done during the process in foot-pounds.
This area is frequently referred to as "the area
under the curve."

Figure 3-5 is a PV diagram of a constant
volume process. Assume that the initial
condition of the gas at the start of the process
is such that the pressure is 2000 psfa and the
volume is 4 cu ft. The gas is heated while its
volume is kept constant until the pressure
increases to 4000 psfa. The process takes place
along the constant volume line from the initial
condition 1 to the final condition 2.

It has been stated that no work is done during
a process unless the volume of the gas changes.
Examination of the PV diagram in Fig. 3-5 will
show that no work is indicated for the constant
volume process. Since a line has only the
dimension of length, there is no area between
the process diagram and the base or volume
axis. Hence, no work is done.
3-25. Constant Temperature Process.
According to Boyle's law, when a gas is com-
pressed or expanded at a constant temperature,
the pressure will vary inversely with the volume.
That is, the pressure increases as the gas is

36 PRINCIPLES OF REFRIGERATION

4500

4000

3500

3000

32500

| 2000

1500

500

^~ Final condition

^ — Initial condition

—

II 1

3 4 5 6 7

Volume (cubic feet)

Fig. 3-5. Pressure-volume diagram of constant
volume process. Since there is no area between the
process diagram and the volume axis, there is no
work done during a constant volume process.

compressed and decreases as the gas is expanded.
Since the gas will do work as it expands, if
the temperature is to remain constant, energy
with which to do the work must be absorbed
from an external source (Fig. 3-26). However,
since the temperature of the gas remains
constant, ail of the energy absorbed by the gas
during the process leaves the gas as work;
none is stored in the gas as an increase in the
internal energy.

When a gas is compressed, work is done on
the gas, and if the gas is not cooled during ihe
compression, the internal energy of the gas will
be increased by an amount equal to the work of
compression. Therefore, if the temperature of
the gas is to remain constant during the com-
pression, the gas must reject to some external
body an amount of heat equal to the amount of
work done upon it during the compression
(Fig. 3-2c).

There is no change in the internal kinetic
energy during a constant temperature process.
Therefore, in Equation 3-12, AK is equal to zero
and the general energy equation for a constant
temperature process may be written

AQ f = AW t (3-21)

3-26. Work of an Isothermal Process. A

PV diagram of an isothermal expansion is

shown in Fig. 3-6. In a constant temperature
process the pressure and volume both change in
accordance with Boyle's law. The path followed
by an isothermal expansion is indicated by
line 1 to 2 and the work of the process in foot-
pounds is represented by the area l-2-V^-V^
The area, 1-2-V^-V^ and therefore the work
of the process, may be calculated by the equa-
tion

fF = P 1 K 1 xln-^

(3-22)

where In = natural logarithm (log to the base e)

Example 3-18. A certain weight of gas
having an initial pressure of 2500 psfa and an
initial volume of 2 cu ft is expanded isothermally
to a volume of 4 cu ft. Determine:

(a) the final pressure of the gas in psfa

(b) the work done in heat energy units.

Solution
(a) By applying Boyle's
law, Equation 3-6,
the final pressure P a

PiVx

Vi
2500 x 2

4
= 1250 psfa

= 2500 x 2
= 2500 x 2

(b) By applying Equa-
tion 3-22, the ex-
ternal work of the

x lnf
x In 2

process in foot-
pounds W

= 2500 x 2
= 3465 ft-Ib

x 0.693

By Applying Equa-
tion 3- 1 8 , the work in
heat energy units Alf

3465
~ 778
= 4.45 Btu

Final
condition

Fig. 3-6. Pressure-volume diagram of constant
temperature process. Crosshatched area represents
the work of the process.

THERMODYNAMIC PROCESSES 37

Example 3-19. A certain weight of gas
having an initial pressure of 12S0 psfa and an
initial volume of 4 cu ft is compressed isotherm-
ally to a volume of 2 cu ft. Determine:

(a) the final pressure in psfa

(b) the work done by the gas in Btu.

Solution

(a) By applying
Boyle's law,
Equation 3-6,
the final pressure

PiVi

1250 x 4

= 2500 psfa

(b) By Applying Equa-
tion 3-22, the ex-

= 2500 x 2 x In |

ternal work of the

= 2500 x 2 x In 0.5

process in foot-

= 2500 x 2 x -0.693

pounds W

= -3465 ft-lb

By applying
Equation 3-18,
the work in heat

-3465
778

energy units AW

= -4.45 Btu

Notice that the process in Example 3-19 is the
exact reverse of that of Example 3-18. Where
the process in Example 3-18 is an expansion,
the process in Example 3-19 is a compression.
Both processes occur between the same two
conditions, except that the initial and final
conditions are reversed. Notice also that
whereas work is done by the gas during the
expansion process, work is done on the gas
during the compression process. But since the
change of condition takes place between the
same limits in both cases, the amount of work
done in each case is the same.
3-27. Heat Transferred during a Constant
Temperature Process. Since there is no
change in the temperature during an isothermal
process, there is no change in the internal
kinetic energy and AK equals zero. According
to Equation 3-21, the heat energy transferred
during a constant temperature process is exactly
equal to the work done in Btu. During an
isothermal expansion heat is transferred to the
gas to supply the energy to do the work that is
done by the gas, whereas during an isothermal
compression heat is transferred from the gas so
that the internal energy of the gas is not in-
creased by the performance of work on the gas.

Example 3-20. Determine the quantity of
heat transferred to the gas during the constant
temperature expansion described in Example
3-18.

Solution. From Example
3-18, AW = 4.45 Btu

Since, in the isothermal
process, AW t equals AQ t , AQ t = 4.45 Btu

Example 3-21. What is the quantity of
heat transferred to the gas during the constant
temperature process described in Example 3-19.

Solution. From
Example 3-19, AW = -4.45 Btu

Since A W t equals AQ t , AQ t = -4.45 Btu

Again, notice that a negative amount of heat
is transferred to the gas, indicating that heat in
this amount is actually given up by the gas
during the process.

3-28. Adiabatic Process. An adiabatic pro-
cess is described as one wherein the gas
changes its condition without absorbing or
rejecting heat, as such, from or to an external
body during the process. Furthermore, the
pressure, volume, and temperature of the gas all
vary during an adiabatic process, none of them
remaining constant.

When a gas expands adiabatically, as in any
other expansion, the gas does external work and
energy is required to do the work. In the
processes previously described, the gas absorbed
the energy to do the work from an external
source. Since, during an adiabatic process, no
heat is absorbed from an external source, the gas
must do the external work at the expense of its
own energy. An adiabatic expansion is always
accompanied by a decrease in the temperature
of the gas as the gas gives up its own internal
energy to do the work (Fig. 3-7).

When a gas is compressed adiabatically,
work is done on the gas by an external body.
The energy of the gas is increased in an amount
equal to the amount of work done, and since no
heat energy is given up by the gas to an external
body during the compression, the heat energy
equivalent of the work done on the gas is set up
as an increase in the internal energy, and the
temperature of the gas increases.

Because no heat, as such, is transferred to or
from the gas during an adiabatic process, AQ a is
always zero and the energy equation for an
adiabatic process is written as follows:

AK a +AW a =0

Therefore,

(3-23)

AW. = -AK. and AK a - -AW a

38 PRINCIPLES OF REFRIGERATION

Final
condition

Fig. 3-7. Pressure-volume diagram of adiabatic
process. An isothermal curve is drawn in for
comparison.

3-29. Work of an Adiabatic Process. The

work of an adiabatic process may be evaluated
by the following equation:

P 1 V 1 -P 2 V 2

W a =

k - 1

(3-24)

where k — the ratio of the specific heats of the
gas in question, CJC V

Example 3-22. A gas having an initial
pressure of 2500 psfa and an initial volume of
2 cu ft is expanded adiabatically to a volume
of 4 cu ft. If the final pressure is 945 psfa,
determine the external work done in heat

energy units.

Solution

Cj, for air
C„ for air

The ratio of the specific
heats, k

= 0.2375 Btu/lb
= 0.169 Btu/lb

0.2375

0.169

= 1.406

Applying Equation 3-24,
the work of adiabatic ex-

(2500 x 2)
- (945 x 4)

pansion in foot-pounds

1406 - 1

w a

5000 - 3780

0.406

1220

0.406

= 3005 ft-lb

Applying Equation 3-18,
the work in heat energy

3005

~ 778

units A W a

= 3.86 Btu

Example 3-23. A gas having an initial
pressure of 945 psfa and an initial volume of
4 cu ft is compressed adiabatically to a volume
of 2 cu ft. If the final pressure of the air is
2500 psfa, how much work is done in heat
energy units?

Solution. From
Example 3-22, k for air

Applying Equation 3-24,
the work done in foot-
pounds W a

1.406

(945 x 4) -
(2500 x 2)
1.406 - 1

3780 - 5000
0.406

-1220

Applying Equation 3-18,
the work in heat energy
units A(C.

0.406

-3005 ft-lb
-3005

778
-3.86 Btu

3-30. Comparison of the Isothermal and
Adiabatic Processes. A comparison of the
isothermal and adiabatic processes is of interest.
Whenever a gas expands, work is done by the
gas, and energy from some source is required to
do the work. In an isothermal expansion, all
of the energy to do the work is supplied to the
gas as heat from an external source. Since the
energy is supplied to the gas from an external
source at exactly the same rate that the gas is
doing work, the internal energy of the gas
neither increases nor decreases and the tempera-
ture of the gas remains constant during the
process. On the other hand, in an adiabatic
expansion there is no transfer of heat to the gas
during the process and all of the work of expan-
sion must be done at the expense of the internal
energy of the gas. Therefore, the internal
energy of the gas is always diminished by an
amount equal to the amount of work done and
the temperature of the gas decreases accordingly.
Consider now isothermal and adiabatic
compression processes. In any compression
process, work is done on the gas by the com-
pressing member, usually a piston, and an
amount of energy equal to the amount of work
done on the gas is transferred to the gas as work.
During an isothermal compression process,
energy is transferred as heat from the gas to an
external sink at exactly the same rate that work
is being done on the gas. Therefore, the internal

THERMODYNAMIC PROCESSES 39

energy of the gas neither increases nor decreases
during the process and the temperature of the
gas remains constant. On the other hand,
during an adiabatic compression, there is no
transfer of energy as heat from the gas to an
external sink. Therefore, an amount of energy
equal to the amount of work done on the gas is
set up in the gas as an increase in the internal
energy, and the temperature of the gas increases
accordingly.

3-34. The Polytropic Process. Perhaps the
simplest way of denning a polytropic process is
by comparison with the isothermal and adia-
batic processes. The isothermal expansion, in
which the energy to do the work of expansion is
supplied entirely from an external source, and
the adiabatic expansion, in which the energy to
do the work of expansion is supplied entirely by
the gas itself, may be thought of as the extreme
limits between which all expansion processes will
fall. Then, any expansion process in which the
energy to do the work of expansion is supplied
partly from an external source and partly from
the gas itself will follow a path which will fall
somewhere between those of the isothermal and
adiabatic processes (Fig. 3-8). Such a process is
known as a polytropic process. If during a
polytropic' expansion most of the energy to do
the work comes from an external source, the
polytropic process will more nearly approach
the isothermal. On the other hand, when the
greater part of the energy to do the external
work comes from the gas itself, the process
more nearly approaches the adiabatic.

This is true also for the compression process.
When a gas loses heat during a compression
process, butnotataratesumcient to maintain the
temperature constant, the compression is poly-
tropic. The greater the loss of heat, the closer
the polytropic process approaches the iso-
thermal.' The smaller the loss of heat, the closer
the polytropic process approaches the adiabatic.
Of course, with no heat loss, the process
becomes adiabatic.

The actual compression of a gas in a compres-
sor will usually very nearly approach adiabatic
compression. This is because the time of com-
pression is normally very short and there is not
sufficient time for any significant amount of heat
to be transferred from the gas through the
cylinder walls to the surroundings. Water
jacketing of the cylinder will usually increase the

Polytropic N > 1 and < K

Fig. 3-8. Pressure-volume diagram of a polytropic
process. Adiabatic and isothermal curves are drawn
in for comparison.

rate of heat rejection and move the path of the
compression closer to the isothermal.
3-32. PVT Relationship during Adiabatic
Processes. Since the temperature, pressure,
and volume all change during an adiabatic
process, they will not vary in accordance with
Charles' and Boyle's laws. The relationship
between the pressure, temperature, and volume
during an adiabatic process may be evaluated by
the following equations :

T t = T x x

* -1)

F (*-l)

p*=Pi

(3-25)

(3-26)

(3-27)

- P '*W

(3-28)

(3-29)

- v, . (3>r

(3-30)

Example 3-24. Air is expanded adiabati-
cally from a volume of 2 cu ft to a volume of

40 PRINCIPLES OF REFRIGERATION

4 cu ft. If the initial pressure of the air is 24,000
psfa, what is the final pressure in psfa?

Solution. From
Table 3-1, k for air

Applying Equation
3-27, the final pressure

,1.406

= 1.406
= 24,000 x

= 24,000 x 0.5 1 * 06
= 24,000 x 0.378
= 9072 psfa

Example 3-25. Air is expanded adia-
batically from a volume of 2 cu ft to a volume
of 4 cu ft. If the initial temperature of the air
is 600° R, what is the final temperature in
degrees Rankine?

Solution. From
Table 3-1, k for air

Applying Equation
3-25, the final tempera-
ture T 2

= 1.406
= 600x

(2)«

.406-1)

= 600x

(4)(1.406-

(2)0.406

(4)0.408

1.325

1.756
= 600 x 0.755
= 453" R

Example 3-26. Air is expanded adia-
batically from an initial pressure of 24,000 psfa
to a final pressure of 9072 psfa. If the initial
temperature is 600° R, what is the final tem-
perature?

Solution. From
Table 3-1, k for
air

Applying
Equation 3-26,
the final tempera-
ture T 9

= 1.406

= 600 x

1.406-1

9072 \ i«o«

/ 9072 \ **
\24,000/

124 000/
= 600 x (0.378)< 0M »»
= 600 x 0.755
= 453° R

3-33. Exponent of Polytropic Expansion
and Compression. The presssure-tempera-
ture-volume relationships for the polytropic
process can be evaluated by Equations 3-25
through 3-30, except that the polytropic ex-
pansion or compression exponent n is substi-
tuted for k. Too, the work of a polytropic
process can be determined by Equation 3-24 if
n is substituted for k.

The exponent n will always have a value
somewhere between 1 and k for the particular

gas ' undergoing the process.* Usually, the
value of « must be determined by actual test of
the machine in which the expansion or com-
pression occurs. In some instances average
values of n for some of the common gases
undergoing changes under more or less standard
conditions are given in tables. If the values of
two properties are known for both initial and
final conditions, the value of n may be calcu-
lated. The following sample equation shows
the relationship:

log (Pj/Pj,)

-^ (3-31)

« =

logC^/^)

Example 3-27." Air, having an initial
pressure of 24,000 psfa and an initial tempera-
ture of 600° R, is expanded polytropically from
a volume of 2 cu ft to a volume of 4 cu ft. If
the exponent of polytropic expansion is 1.2,
determine:

(a) The weight of the air in pounds

(6) The final pressure in psfa

(d) The work done by the gas in Btu

(e) The increase in internal energy
(/) The heat transferred to the gas.

Solution
(a) From Table 3-1,

R for air

Rearranging and
applying Equation
3-10, the weight of
air M

= 53.3

RT
_ 24,000 x 2

53.3 x 600
= 1.5 lb

* The value of n depends upon the specific heat
of the gas during the process. Since the specific
heat may take any value, it follows that theoretically
n may have any value. In actual machines, however,
n will nearly always have some value between 1
and A:.

Broadly- defined, a polytropic process is any
process during which the specific heat remains
constant. By this definition, all five processes
discussed in this chapter are polytropic processes.
It is general practice today to restrict the term
polytropic to mean only those processes which follow
a path falling between those of the isothermal and
adiabatic processes. The exponents of isothermal
and adiabatic expansions or compressions are 1
and k, respectively. Hence, the value of n for the
polytropic process must fall between 1 and k. The
closer the polytropic process approaches the
adiabatic, the closer n will approach k.

(b) Applying Equation
3-27, the final

- 24,000 xg)"

pressure P a

= 24,000 x (0.SY*

- 24,000 x 0.435

= 10,440 psfa

(2)U.»-i)

— 600 x

3-25, the tempera-

""" * (4)<i.«-i)

ture T t

-«0x3K

(4) o.»

,„ 1.149

= 600 x — —

1.32

= 522°R

(d) Applying Equation

(24,000 x 2)

3-24, the work

- (4 x 10,440)

done W

1.2 - 1

_ 48,000 - 41,760

0.2

6240

0.2

= 31,200 ft-lb

Applying Equation

31,200

3-18, the work in

778

BtuAW

= 40.10 Btu

(e) From Table 3-1,

= 0.169 Btu/lb

C„ for air

Applying Equation

= 1.5 x 0.169 x

3-15, the increase

(522 - 600)

in internal energy

=■ 1.5 x 0.169 x

(-78)

= -19.77 Btu

(/) Applying Equation

= AJSr + AfF

3-9, the heat

energy trans-

= -19.77 + 40.10

ferred to the gas AQ

= 20.33 Btu

Notice in Example 3-27 that the work done
by the air in the polytropic expansion is equiva-
lent to 40.10 Btu. Of this amount, 20.33 Btu
is supplied from an external source, whereas the
other portion, 19.77 Btu, is supplied by the gas
itself, thereby reducing the internal kinetic
energy by this amount.

PROBLEMS

1. Three pounds of air occupy a volume of
24 cu ft. Determine:

(a) The density of the air. Arts. 0.125 lb/cu ft

(b) The specific volume. Am. 8 cu ft/lb

THERMODYNAMIC PROCESSES 41

2. The volume of a certain weight of air is kept
constant while the temperature of the air is
increased from 55° F to 100° F. If the initial
pressure is 25 psig, what is the final pressure of
the air in psig ? Am. 28 .47 psig

3. A certain weight of air confined in a con-
tainer is cooled from 150° F to 70° F. If the
initial pressure of the air is 36.3 psig, what is the
final pressure of the air in psig ? Am. 29.6 psig

4. One pound of air at atmospheric pressure
has a volume of 13.34 cu ft at a temperature
of 70° F. If the air is passed across a heat
exchanger and is heated to a temperature of
150° F while its pressure is kept constant, what
is the final volume of the air ? Ans. 15.35 cu ft

5. A cylinder of oxygen has a volume of 5 cu ft.
A gage on the cylinder reads 2200 psi. If the
temperature of the oxygen is 85° F, what is the
weight of the oxygen in the cylinder?

Am. 60.6 lb

6. In Problem 4, determine:

(a) The work done by the air during the
heating. Am. 4254.8 ft-lb or 5.47 Btu

(b) The increase in the internal kinetic energy.

Am. 13.52 Btu

Am. 19 Btu

7. A certain weight of air having an initial
volume of 0.1334 cu ft and an initial tempera-
ture of 70° F is drawn into the suction side of
an air compressor. If the air enters the cylinder
at standard atmospheric pressure and is com-
pressed isothermally to a final pressure of 150
psia, determine:

(a) The weight of air in the cylinder at the
start of the compression stroke.

Am. 0.01 lb

(b) The final temperature of the air in degrees
Rankine. Am. 530° R

(d) The work of compression in Btu.

Am. 0.843 Btu

(e) The increase or decrease in internal energy.

Am. None

(/) The energy transferred to the gas during

the compression. Am. —0.843 Btu

8. Assume that the air in Problem 7 is com-
pressed adiabatically rather than isothermally.
Compute:

(a) The final temperature of the air in degrees
Rankine. Am. 1038° R

(b) The volume of the air at the end of the
compression stroke. Am. 0.0256 cu ft

Am. 0.86 Btu

42 PRINCIPLES OF REFRIGERATION

(d) The increase in the internal kinetic energy.

Arts. 0.86 Btu
(c) The heat energy transferred to or from
the gas during the compression.

Ans. None

9. Assuming that the air in Problem 7 is com-
pressed polytropically rather than isothermally.
If n equals 1.2, compute:
(«) The final temperature of the air in degrees
Rankine. Ans. 119 A" R

(b) The volume of the air at the end of the
compression stroke. Ans. 0.0192 cu ft

Ans. 0.8S Btu

(d) The decrease in the internal kinetic energy.

Ans. 0.45 Btu

(e) The heat energy transferred from the gas
during the compression. Ans. 0.40 Btu

10. Compare the results of Problems 7, 8,
and 9.

Saturated and

Superheated

Vapors

4-1. Saturation Temperature. When the
temperature of a liquid is raised to a point such
that any additional heat added to the liquid will
cause a part of the liquid to vaporize, the
liquid is said to be saturated. Such a liquid is
known as a saturated liquid and the tempera-
ture of the liquid at that condition is called the
saturation temperature (Sections 2-31 and 2-32).
4-2. Saturated Vapor. The vapor ensuing
from a vaporizing liquid is called a saturated
vapor as long as the temperature and pressure
of the vapor are the same as those of the satur-
ated liquid from which it came. A saturated
vapor may be described also as a vapor at a
temperature such that any further cooling of
the vapor will cause a portion of the vapor to
condense and thereby resume the molecular
structure of the liquid state. It is important
to understand that the saturation temperature
of the liquid (the temperature at which the liquid
will vaporize if heat is applied) and the satura-
tion temperature of the vapor (the temperature
at which the vapor will condense if heat is
removed) are the same for any given pressure
and that the liquid cannot exist as a liquid at any
temperature above its saturation temperature,
whereas a vapor cannot exist as a vapor at any
temperature below its saturation temperature.*

* Under certain conditions it is possible to "super-
cool" water vapor momentarily below its saturation
temperature. However, this is a very unstable
condition and cannot be maintained except momen-
tarily.

For example, in Fig. 4-1 , the water in the heated
vessel is saturated and is vaporizing at 212° F as
the latent heat of vaporization is supplied by the
burner. The water vapor (steam) rising from
the water is saturated and remains at the satura-
tion temperature (212° F) until it reaches the
condenser. As the saturated vapOr gives up
heat to the cooler water in the condenser, it
condenses back into the liquid state. Since
condensation occurs at a constant temperature,
the water resulting from the condensing vapor
is also at 212° F. The latent heat of vapor-
ization, absorbed as the water vaporizes into
steam, is given up by the steam as the steam
condenses back into water.
4-3. Superheated Vapor. A vapor at any
temperature above its saturation temperature
is a superheated vapor (Section 2-34). If, after
vaporization, a vapor is heated so that its
temperature is raised above that of the vapor-
izing liquid, the vapor is said to be superheated.
In order to superheat a vapor it is necessary to
separate the vapor from the vaporizing liquid as
shown in Fig. 4-2. As long as the vapor remains
in contact with the liquid it will be saturated.
This is because any heat added to a liquid-
vapor mixture will merely vaporize more liquid
and no superheating will occur.

Before a superheated vapor can be condensed,
the vapor must be de-superheated, that is, the
vapor must first be cooled to its saturation
temperature. Heat removed from a super-
heated vapor will cause the temperature of the
vapor to decrease until the saturation tempera-
ture is reached. At this point, any further
removal of heat will cause a part of the vapor to
condense.

4-4. Subcooled Liquid. If, after condensa-
tion, a liquid is cooled so that its temperature
•is reduced below the saturation temperature, the
liquid is said to be subcooled. Thus, a liquid at
any temperature below the saturation tempera-
ture and above the fusion point is a subcooled
liquid.

4-5. The Effect of Pressure on the Satura-
tion Temperature. The saturation tempera-
ture of a liquid or a vapor varies with the
pressure. Increasing the pressure raises the
saturation temperature and decreasing the pres-
sure lowers the saturation temperature. For
example, the saturation temperature of water at
atmospheric pressure (0 psig or 14.7 psia) is

44 PRINCIPLES OF REFRIGERATION

212° F. If the pressure over the water is in-
creased from psig to 5.3 psig (20 psiaX the
saturation temperature of the water increases
from 212° F to 228° F. On the Other hand, if the
pressure over the water is reduced from 14.7
psia to 10 psia, the new saturation temperature
of the water will be 193.2° F. Figure 4-3 is a

the water at atmospheric pressure is 212° F,
the temperature of the water will rise as the
water is heated until it reaches 212° F. At this
point, if the heating is continued, the water will
begin to vaporize. Soon the space above the
water will be filled with billions and billions of
water vapor molecules darting about at high

Condenser
water out""^

Saturated steam
at212"F -*

Heat added

Steam gives up heat to

, cold water in condenser

/ and condenses into water

.{ ^Con densed steam leaving
condenser-212°F

Cold water in

Fig. 4VI. Saturated vapor.

graphical representation of the relationship
between the pressure and the saturation tempera-
ture of water.

To illustrate the effect of pressure on the
saturation temperature of a liquid, assume that
water is confined in a closed vessel which is
equipped with a throttling valve at the top
(Fig. 4-4a). A compound gage is used to deter-
mine the pressure exerted in the vessel and two
thermometers are installed so that one records
the temperature of the water and the other the
temperature of the vapor over the water. With
the throttling valve wide open, the pressure
exerted over the water is atmospheric (0 psig or
14.7 psia). Since the saturation temperature of

velocities. Some of the vapor molecules will fall
back into the water to become liquid molecules
again, whereas others will escape through the
opening to the outside and be carried away by
air currents. If the opening at the top of the
vessel is of sufficient size to allow the vapor to
escape freely, the vapor will leave the vessel at
the same rate that the liquid is vaporizing. That
is, the number of molecules which are leaving
the liquid to become vapor molecules will be
exactly equal to the number of vapor molecules
which are leaving the space, either by escaping
to the outside or by falling back into the liquid.
Thus, the number of vapor molecules and the
density of the vapor above the liquid will

SATURATED AND SUPERHEATED VAPORS 45

Fig. 4-2. Superheated vapor.

« r- : iOiA-.t!j>.^. -^j^~ -± Jt.i-^x4j— ^ ^.j ^Ls ^Steam superheated

in superheater

Saturated steam

tint

Heat added

remain constant and the pressure exerted by the
vapor will be equal to that of the atmosphere
outside of the vessel.

Under this condition the water vapor ensuing
from the vaporizing liquid will be saturated,
that is, its temperature and pressure will be the
same as that of the water, 212° F and 14.7 psia.
The density of the water vapor at that tempera-
ture and pressure will be 0.0373 lb/cu ft and its
specific volume will be 1 /0.0373 or 26.8 ft*/lb.

Regardless of the rate at which the liquid is
vaporizing, as long as the vapor is allowed to
escape freely to the outside so that the pressure
and density of the vapor over the liquid does
not change, the liquid will continue to vaporize
at 212° F.

/T7-X

TfT"/ 5 "

H eat Superheated

added

steam

Suppose that the throttling valve is partially
closed so that the escape of the vapor from the
vessel is impeded somewhat. For a time the
equilibrium will be disturbed in that the vapor
will not be leaving the vessel at the same rate the
liquid is vaporizing. The number of vapor
molecules in the space above the liquid will
increase, thereby increasing the density and the
pressure of the vapor over the liquid and raising
the saturation temperature.

If it is assumed that the pressure of the vapor
increases to 5.3 psig (20 psia) before equilibrium
is again established, that is, before the rate at
which the vapor is escaping to the outside is
exactly equal to the rate at which the liquid is
vaporizing, the saturation temperature will be

DUU

500
-.400

re
'3»
o.

Is 300
a. 200

100
14.7

Fig. 4-3. Variation in the
saturation temperature of
water with changes in pressure.

50 100 150 200| 250 300 350 400 450 500
212
Saturation temperature (*F)

46 PRINCIPLES OF REFRIGERATION

xtt v' Valve wit,e °P en

« /S^ Steam <„'•>
Density-0.0373 Ib/cu ft

: Z z _Water_ H>: : >^^

Valve partially closed

20 psia

►212'F

^i'-'i- Steam
Density-0.0498 Ib/cu ft

>228'F

(a)

<b)

Fig. 4-4.

228° F, the density of the vapor will be 0.0498
lb/cu ft, and 1 lb of vapor will occupy a volume
of 20.08 cu ft. This condition is illustrated in
Fig. 4-46.

By comparing the condition of the vapor in
Fig. 4-4A with that of the vapor in Fig. 4-4a, it
will be noted that the density of the vapor is
greater at the higher pressure and saturation
temperature. Furthermore, it is evident that the
pressure and the saturation temperature of
liquid or vapor can be controlled by regulating
the rate at which the vapor escapes from over
the liquid.

In Fig. 4-4«, the rate of vaporization will have
little or no effect on the pressure and saturation
temperature because the vapor is allowed to
escape freely so that the density and pressure of
the vapor over the liquid will neither increase
nor decrease as the rate of vaporization is
changed. On the other hand, in Fig. 4-46, any
increase in the rate of vaporization will cause an
increase in the density and pressure of the vapor
and result in an increase in the saturation
temperature. The reason is that any increase
in the rate of vaporization will necessitate the
escape of a greater quantity of vapor in a given
length of time. Since the size of the vapor outlet
is fixed by the throttling action of the valve, the
pressure of the vapor in the vessel will increase
until the pressure difference between the inside

and outside of the vessel is sufficient to allow
the vapor to escape at a rate equal to that at
which the liquid is Vaporizing. The increase in
pressure, of course, results in an increase in the
saturation temperature and in the density of the
vapor. Likewise, any decrease in the rate of
vaporization will have the opposite effect. The
pressure and density of the vapor over the liquid
will decrease and the saturation temperature
will be lower.

Assume now that the throttling valve on the
container is again opened completely, as in
Fig. 4-4a, so that the vapor is allowed to escape
freely and unimpeded from over the liquid. The
density and pressure of the vapor will decrease
until the pressure of the vapor is again equal to
that of the atmosphere outside of the container.
Since the saturation temperature of water at
atmospheric pressure is 212° F and since a liquid
cannot exist as a liquid at any temperature above
its saturation temperature corresponding to its
pressure, it is evident that the water must cool
itself from 228° F to 212° F at the instant that
the pressure drops from 20 psia to atmospheric
pressure. To accomplish this cooling, a portion
of the liquid will "flash" into a vapor. The
latent heat necessary to vaporize the portion of
the liquid that flashes into the vapor state is
supplied by the mass of the liquid and, as a
result of supplying the vaporizing heat, the

SATURATED AND SUPERHEATED VAPORS 47

temperature of the mass of the liquid will be
reduced to the new saturation temperature.
Enough of the liquid will vaporize to provide
the required amount of cooling.
4-6. Vaporization. The vaporization of a
liquid may occur in two ways: (1) by evapora-
tion and (2) by ebullition or "boiling." The
vaporization of a liquid by evaporation occurs
only at the free surface of the liquid and may
take place at any temperature below the satura-
tion temperature. On the other hand, ebullition
or boiling takes place both at the free surface
and within the body of the liquid and can occur
only at the saturation temperature. Up to this
point, only ebullition or boiling has been
considered.

4-7. Evaporation. Evaporation is taking
place continually and the fact that water
evaporates from lakes, rivers, ponds, clothes,
etc., is sufficient evidence that evaporation can
and does occur at temperatures below the
saturation temperature. Any liquid open to the
atmosphere, regardless of its temperature, will
gradually evaporate and be diffused into the
air.

The vaporization of liquids at temperatures
below their saturation temperature can be
explained in this manner. The molecules of a
liquid are in constant and rapid motion, their
velocities being determined by the temperature
of the liquid. In the course of their movements
the molecules are continually colliding with one
another and, as a result of these impacts, some
of the molecules of the liquid momentarily
attain velocities much higher than the average
velocity of the other molecules of the mass.
Thus, their energy is much greater than the
average energy of the mass. If this occurs
within the body of the liquid, the high velocity
molecules quickly lose their extra energy in
subsequent collisions with other molecules.
However, if the molecules attaining the higher
than normal velocities are near the surface, they
may project themselves from the surface of the
liquid and escape into the air to become vapor
molecules. (Fig. 4-5). The molecules so
escaping from the liquid are diffused throughout
the air. They occupy the relatively large spaces
which exist between the molecules of the air and
become a part of the atmospheric air.
4-8. Rate of Vaporization. For any given
temperature, some liquids will evaporate faster

High energy molecules
escape from surface
S" of liquid and become
f vapor molecules

Water

Fig. 4-5. Evaporation from surface of a liquid.

than others. Liquids havingthelowest "boiling"
points, that is, the lowest saturation temperature
for a given pressure, evaporate at the highest
rate. However, for any particular liquid, the
rate of vaporization varies with a number of
factors. In general, the rate of vaporization
increases as the temperature of the liquid
increases and as the pressure over the liquid
decreases. Evaporation increases also with the
amount of exposed surface. Furthermore, it
will be shown later that the rate of evaporation
is dependent on the degree of saturation of the
vapor which is always adjacent to and above
the liquid.

4-9. The Cooling Effect of Evaporation.
Since it is the higher velocity molecules (those
having the most energy) which escape from the
surface of an evaporating liquid, it follows that
the average energy of the mass is thereby
reduced and the temperature of the mass
lowered. Whenever any portion of a liquid
vaporizes, an amount of heat equal to the latent
heat of vaporization must be absorbed by that
portion, either from the mass of the liquid,
from the surrounding air, or from adjacent
objects. Thus, the energy and temperature of
the mass are reduced as it supplies the latent
heat of vaporization to that portion of the
liquid which vaporizes. The temperature of the
mass is reduced to a point slightly below that of
the surrounding media and the temperature
difference so established causes heat to flow
from the surrounding media into the mass of the
liquid. The energy lost by the mass during
vaporization is thereby replenished and evapora-
tion becomes a continuous process as long as

48 PRINCIPLES OF REFRIGERATION

Molecules escaping from
surface are carried away
by air so that evaporation
is continuous

Water temperature
slightly below air >.
temperature >

Vapor pressure
0.74 in. Hg
Absolute density
0.001148 Ib/cu ft

Molecules cannot
escape-fall back
into liquid to replace
those leaving

Saturated vapor

70° F
Saturated liquid

(a)

(b)

Vapor pressure-
1.13 in. Hg
Absolute density-
0.001570 Ib/cu ft

Saturated vapor

Vapor pressure—
0.52 in. Hg
Absolute density—
60' F / 0.000828

Saturated vapor

60° F

Saturated liquid

(c)

(d)

Fig. 4-6.

any of the liquid remains. The vapor resulting
from evaporation is diffused into and carried
away by the air.

4-10. Confined Liquid-Vapor Mixtures.
When a vapor is confined in a container with a
portion of its own liquid, both the vapor and the
liquid will be saturated. To illustrate, assume
that an open container is partially filled with
water and is stored where the ambient tempera-
ture is 70° F (Fig. 4-6a). The water will be
evaporating at 70° F and, as described in the
previous section, the vapor molecules leaving
the liquid will be diffused into the surrounding
air so that evaporation will continue until all
of the liquid is evaporated. However, if a
tightly fitting cover is placed over the container,
the vapor molecules will be unable to escape to
the outside and they will collect above theliquid.
Soon the space above the liquid will be so filled
with vapor molecules that there will be as many
molecules falling back into the liquid as there
are leaving the liquid. A condition of equi-
librium will be attained, the vapor will be

saturated, and no further evaporation will
occur. The energy of the liquid will be increased
by the vapor molecules which are returning to
the liquid in exactly the same amount that it is
diminished by the molecules that are leaving.
Since no further cooling will take place by
evaporation, the liquid will assume the tempera-
ture of the surrounding air and' heat transfer
will cease. (Fig. 4-66).

If, at this point, the ambient temperature
rises to, say, 80° F, heat transfer will again take
place between the surrounding air and the
liquid. The temperature and average molecular
velocity of the liquid will be increased and
evaporation will be resumed. The number of
molecules leaving the liquid will again be
greater than the number returning and the
density and pressure of the vapor above the
liquid will be increased. As the density and
pressure of the vapor increase, the saturation
temperature of the liquid increases. Eventually,
when the saturation temperature reaches 80° F
and is equal to the ambient temperature, no

further heat transfer will occur and evaporation
will cease. Equilibrium will have again been
established. The density and pressure of the
vapor will be greater than before, the saturation
temperature of the liquid-vapor mixture will be
higher, and there will be more vapor and less
liquid in the container than previously (Fig.
4-6c).

Suppose now that the ambient temperature
falls to 60° F. When this occurs, heat will flow
from the 80° F liquid-vapor mixture to the
cooler surrounding air. As the liquid-vapor
mixture loses heat to the surrounding air, its
temperature and average molecular velocity will
be decreased and many of the vapor molecules,
lacking sufficient energy to remain in the vapor
state, will fall back into the liquid and resume
the molecular arrangement of the liquid state;
that is, a part of the vapor will condense. The
density and pressure of the vapor will be
diminished and the saturation temperature of
the mixture will be reduced. When the satura-
tion temperature of the mixture falls to 60° F
it will be the same as the ambient temperature
and no further heat flow will occur. Equilibrium
will have been established and the number of
molecules re-entering the liquid will exactly
equal those which are leaving. At this new
condition, the density and pressure of the vapor
will be less than before, the saturation tempera-
ture will be lower, and since a part of the vapor
condensed into liquid, there will be more liquid
and less vapor comprising the mixture than at
the previous condition (Fig. 4-6d).
4-11. Sublimation. It is possible for a
substance to go directly from the solid state
to the vapor state without apparently passing
through the liquid state. Any solid substance
will sublime at any temperature below its
fusion temperature. Sublimation takes place
in a manner similar to evaporation, although
much slower, in that the higher velocity
molecules near the surface escape from the mass
into the surrounding air and become vapor
molecules. One of the most familiar examples
of sublimation is that of solid C0 2 (dry ice),
which, at normal temperatures and pressures,
sublimes directly from the solid to the vapor
state. Damp wash frozen on the line in the
winter time will sublime dry. During freezing
weather ice and snow will sublime from streets
and sidewalks, etc.

SATURATED AND SUPERHEATED VAPORS 49

4-12. Condensation. Condensation of a
vapor may be accomplished in several ways: (1)
by extracting heat from a saturated vapor, (2)
by compressing the vapor while its temperature
remains constant, or (3) by some combination
of these two methods.

4-13. Condensing by Extracting Heat from
a Saturated Vapor. A saturated vapor has
been previously described as one at a condition
such that any further cooling will cause a part
of the vapor to condense. This is because a
vapor cannot exist as a vapor at any tempera-
ture below its saturation temperature. When
the vapor is cooled, the vapor molecules cannot
maintain sufficient energy and velocity to over-
come the attractive forces of one another and
remain as vapor molecules. Some of the
molecules, overcome by the attractive forces,
will revert to the molecular structure of the
liquid state. When condensation occurs while
the vapor is confined so that the volume remains
constant, the density and pressure of the vapor
will decrease so that there is a decrease in the
saturation temperature. If, as in a vapor
condenser (Fig. 4-1), more vapor is entering
the vessel as the vapor condenses and drains
from the vessel as a liquid, the density, pressure,
and saturation temperature of the vapor will
remain constant and condensation will continue
as long as heat is continuously extracted from
the vapor.

4-14. Condensing by Increasing the Pres-
sure at a Constant Temperature. When a
vapor is compressed at a constant temperature,
its volume diminishes and the density of the
vapor increases as the molecules of the vapor
are forced into a smaller volume. The satura-
tion temperature of the vapor increases as the
pressure increases until a point is reached where
the saturation temperature of the vapor is equal
to the actual temperature of the vapor. When
this occurs, the density of the vapor will be at a
maximum value for that condition, and any
further compression will cause a part of the
vapor to assume the more restrained molecular
structure of the liquid state. Thereafter,
condensation will continue as long as com-
pression continues so that the density and
pressure of the remaining vapor cannot be
further increased. If the temperature of the
vapor is to remain constant, heat must be
removed from the vapor during the compression

50 PRINCIPLES OF REFRIGERATION

(Section 3-25). If heat is not removed from the
vapor, the temperature of the vapor will
increase and condensation will not occur.

A careful analysis of Sections 4-13 and 4-14
will show that in either case the vapor is
brought to a saturated condition before conden-
sation begins and that heat is removed from
the vapor in order to bring about condensation.
Furthermore, the vapor is saturated in each
case only when the saturation temperature and
the actual temperature of the vapor are the
same.

In Section 4-13, heat is removed from the
vapor at a constant pressure until the tempera-
ture of the vapor falls to the saturation tempera-
ture corresponding to its pressure, whereupon
the continued removal of heat causes a part of
the vapor to condense.

In Section 4-14, the pressure of the vapor
is increased while the temperature of the vapor
remains constant until the saturation tempera-
ture of the vapor corresponding to the increased
pressure is equal to the actual temperature of
the vapor. In both cases, since the vapor must
give up the latent heat of vaporization in order
to condense, heat must be removed from the
vapor.

4-15. Critical Temperature. The tempera-
ture of a gas may be raised to a point such
that it cannot become saturated regardless of
the amount of pressure applied. The critical
temperature of any gas is the highest tempera-
ture the gas can have and still be condensible
, by the application of pressure. The critical
temperature is different for every gas. Some
gases have high critical temperatures while the
^critical" temperatures of others are relatively
low. For example, the critical temperature of
water vapor is 706° F, whereas the critical
temperature of air is approximately -225° F.
4-16. Critical Pressure. Critical pressure is
the lowest pressure at which a substance can
exist in the liquid state at its critical tempera-
ture; ihit is, it is the saturation pressure at the
critical temperature.

44J. Igriportant Properties of Gases and
Vapors. Although a gas or vapor has many
properties, only six are of particular importance
in the study of refrigeration. These are pressure,
temperature, volume, enthalpy, internal energy,
and entropy. Pressure, temperature, and
volume are called measurable properties be-

cause they can actually be measured. Enthalpy,
internal energy, and entropy cannot be measured.
They must be calculated and are therefore
known as calculated properties.

Pressure, temperature, volume, and internal
energy have already been discussed to some
extent. A discussion of enthalpy and entropy
follows.

4-18. Enthalpy. Enthalpy is a calculated
property of matter which is sometimes loosely
defined as "total heat content." More specifi-
cally, the enthalpy H of a given mass of material
at any specified thermodynamic condition is an
expression of the total heat which must be
transferred to the material to bring the material
to the specified condition from some initial
condition arbitrarily taken as the zero point of
enthalpy.

Whereas the total enthalpy H represents the
enthalpy of M pounds, the specific enthalpy
h is the enthalpy of 1 lb. Since it is usually the
specific enthalpy rather than the total enthalpy
which is of interest, hereafter in this text the
term enthalpy shall be used to mean specific
enthalpy, h, the enthalpy of 1 lb.

Since little is known about the specific heat or
the other properties of materials at low tempera-
tures, it is not possible to determine absolute
values for the calculated properties. For this
reason, values for the calculated properties
must be determined from some arbitrarily
selected zero point rather than from absolute
zero.* For example, the zero point of enthalpy
for water and its vapor, steam, is taken as
water at 32 a F under atmospheric pressure.
The enthalpy of 1 lb of water at 60° F then is the
total amount of heat which must be transferred
to the water in order to raise the temperature of
the water from 32° F to 60° F. According to
Equation 2-9, this is 28 Btu (1 x 1 x 28).
Hence, based on the assumption that the
enthalpy of water is zero at 32° F, the enthalpy
of water at 60° F is 28 Btu/lb.

Mathematically, enthalpy is defined as

(4-1)

* Since it is required to know the change in the
enthalpy of the working fluid during a process,
rather than the absolute enthalpy at some par-
ticular condition, the fact that absolute enthalpy
cannot be calculated is of little consequence.

SATURATED AND SUPERHEATED VAPORS 51

Standard atmospheric pressure
2105 psfa (14.6% x 144)

Fig. 4-7. Pressure-volume
diagram showing the external
work done by fluid expansion
as I lb of water is vaporized
at atmospheric pressur
approximately 59,000 ft-lb.

« P

i s- Final condition

Specific volume of
1 lb of steam at 212°
26.8 cu ft/lb

Volume cubic feet

where h = the specific enthalpy in Btu/lb

u = the specific internal energy in Btu/lb
P = the pressure in psfa
v = the specific volume in cubic feet
J = the mechanical energy equivalent
It has been demonstrated (Section 3-12) that
all the heat transferred to a fluid is not neces-
sarily stored in the fluid as an increase in the
internal energy of the fluid. In many cases,
some part or all or the heat transferred to the
fluid passes through the fluid and leaves the
fluid as work. In Equation 4-1, that part of the
transferred energy which is stored in the fluid
as an increase in the internal energy is repre-
sented by the term u, whereas that part of the
transferred heat which leaves the fluid as work
is represented by the term PvjJ. Notice that,
although the energy represented by the term
PvjJ, does not increase the internal energy of the
fluid and is not stored in the fluid, it neverthe-
less represents energy which must be trans-
ferred to the fluid in order to bring the fluid
to the specified condition from the initial
condition at the arbitrarily selected zero point of
enthalpy. Furthermore, even though the
external work energy is not stored in the fluid,
it must pass back through the fluid and be given
up by the fluid as the fluid returns to the initial
condition.

Consider, for example, the vaporization of 1
lb of water into steam at 212° F under atmos-
pheric pressure. The volume of 1 lb of water at
212° F is 0.01670 cu ft whereas the volume of 1
lb of steam at 212° F is 26.82 cu ft. Hence, the
fluid expands from a volume of 0.0167 cu ft to a
volume of 26.82 cu ft during the vaporization
thereby doing work in expanding against the
pressure of the atmosphere.

The enthalpy of vaporization (latent heat of
vaporization) of water at 212° F is 970.4 Btu. Of
this amount, only 897.6 Btu actually increases
the internal energy and represents energy in
storage in the vapor. The other 72.8 Btu leaves
the vapor as the work of expansion and is
represented by the term PvjJ. A PV diagram of
the vaporization process is shown in Fig. 4-7.
4-19. Entropy. Entropy, like enthalpy, is a
calculated property of matter. The entropy S
of a given mass of material at any specified
condition is an expression of the total heat
transferred to the material per degree of abso-
lute temperature to bring the material to that
condition from some initial condition taken as
the zero of entropy.

Since it is not possible to calculate the
absolute value of entropy, entropy values, like
those of enthalpy, are based on an arbitrarily
selected zero point. The zero points of entropy
and enthalpy are the same for any one fluid.
Hence, for water and its vapor, steam, the zero
point of entropy is taken as water at 32° F.

Again, as in the case of enthalpy, it is the
specific entropy s rather than the total entropy
S which is useful. Therefore, in this book, the
term entropy shall be used to mean specific
entropy s rather than the total entropy S.

It has been shown (Section 3-22) that the
mechanical energy or work of a process can be
expressed as the product of the change in
volume and the average absolute pressure.
Likewise, it is often convenient to express the
heat energy transferred during a process as the
product of two factors. The concept of entropy
makes this possible. The heat energy trans-
ferred during a process can be expressed as the
product of the change in entropy and the

52 PRINCIPLES OF REFRIGERATION

■672*R(212'F + 460)

Fig. 44

Specific entropy of
saturated steam at 212° F

g]/ 1.7566 Btu/lb/°R

Entropy (Btu/lb/'R)

average absolute temperature. * Mathematically
the relationship is expressed by the following
equations:

AQ = As x T m (4-2)

a A <2

* m

T = — —
lm As

(4-3)
(4-4)

where AQ = the heat energy transferred in Btu
As = the change in entropy in Btu per

pound per ° R
T m = the average absolute temperature
in°R

On a pressure-volume diagram (Fig. 4-7), the
"area under the curve," which is the product
of the change in volume and the average abso-
lute pressure, represents the work of the process.
Similarly, on a temperature-entropy diagram
(temperature plotted against entropy), the
"area under the curve," which is the product
of the entropy change and the average absolute
temperature, represents the heat transferred
during the process (Fig. 4-8).

Although the mathematical treatment of
entropy is not required in the study of refrigera-
tion and is beyond the scope of this book, it is
important to note that according to Equation
4-2 the entropy changes only when heat is
transferred during the process. If there is no
heat energy transfer, there is no change in the
entropy. The heat energy transfer may occur
either to or from an external source or sink or it

* The average absolute temperature is not merely
the mean of the initial and final temperatures of the
process, but is the average of all of the absolute
temperatures through which the process passes.

may take place entirely within the fluid itself
as a result of internal friction. However, the
entropy of a fluid is not affected by external
work done either by or on the fluid. Thus in a
frictionless (occurring without either internal
or external friction), adiabatic (no heat transfer
to or from an external body) compression, as in
the ideal compression of the refrigerant vapor
in a refrigeration compressor, the entropy of the
fluid will remain the same or constant.
4-20. Vapor Tables. It has been stated
previously that a vapor does not approach the
condition of an ideal gas because of the inter-
molecular forces which exist between the
molecules of the vapor. Therefore, internal
friction is present whenever a vapor undergoes a
change of condition so that the various proper-
ties of a vapor at the different conditions cannot
be determined by applying the laws of ideal

The properties of vapors at various conditions
have been determined by experiment for all
common vapors and these data are published in
the form of tables. Separate tables are used for
saturated and superheated vapors.
4-21. Saturated Vapor Tables. Saturated
vapor tables (Fig. 4-9) deal only with saturated
liquids and vapors, and usually give values for
the following properties: (1) temperature, (2)
pressure, (3) specific volume, (4) enthalpy
(specific), and (5) entropy (specific). Normally,
the temperature in degrees Fahrenheit is listed
in the extreme left-hand column. The pressure
is given in the second and third columns,
followed by the specific volume in cubic feet
for both the liquid and the vapor in the fourth
and fifth columns, respectively. Some tables
list the density in addition to or in place of the

SATURATED AND SUPERHEATED VAPORS
Properties of Saturated Steam

Absolute Pressure

Specific Volume

Enthalpy

Entropy

Temp.,

Sat.

"F,

(1)

Psi,

P
(2)

In. Hg,

P
(3)

liquid,
(4)

Evap.,
(5)

vapor,

(6)

liquid,
h,
(7)«

Evap.,

hfg

(8)

vapor,

(9)

liquid,

(10)

Evap.,

St,
(11)

vapor,

(12)

200

11.526

23.467

0.01663

33.62

33.64

167.99

977.9

1145.9

0.2938

1.4824

1.7762

202

12.011

24.455

0.01665

32.35

32.37

170.00

976.6

1146.6

0.2969

1.4760

1.7729

204

12.512

25.475

0.01666

31.14

31.15

172.02

975.4

1147.4

0.2999

1.4697

1.7696

206

13.031

26.531

0.01667

29.97

29.99

174.03

974.2

1148.2

0.3029

1.4634

1,7663

208

13.568

27.625

0.01669

28.86

28.88

176.04

972.9

1148.9

0.3059

1.4571

1.7630

210

14.123

28.755

0.01670

27.80

27.82

178.05

971.6

1149.7

0.3090

1.4508

1.7598

212

14.696

29.922

0.01672

26.78

26.80

180.07

970.3

1150.4

0.3120

1.4446

1.7566

214

15.289

31.129

0.01673

25.81

25.83

182.08

969.0

1151.1

0.3149

1.4385

1.7534

216

15.901

32.375

0.01674

24.88

24.90

184.10

967.8

1151.9

0.3179

1.4323

1.7502

218

16.533

33.662

0.01676

23.99

24.01

186.11

966.5

1152.6

0.3209

1.4262

1.7471

220

17.186

34.992

0.01677

23.13

23.15

188.13

965.2

1153.4

0.3239

1.4201

1.7440

222

17.861

36.365

0.01679

22.31

22.33

190.15

963.9

1154.1

0.3268

1.4141

1.7409

224

18.557

37.782

0.01680

21.53

21.55

192.17

962.6

1154.8

0.3298

1.4080

1.7378

226

19.275
C2O.O160

20.780

39.244

0.01682

20.78

20.79

194.18

961.3

1155.5

0.3328

1.4020

1.7348

228

40.753

0.01683

20.06

20.07

196.20

960.1

1156.3

0.3357

1.3961

1.7318

230

42.308

0.01684

19.365

19.382

198.23

958.8

1157.0

0.3387

1.3901

1.7288

240

24.969

50.837

0.01692

16.306

16.323

208.34

952.2

1160.5

0.3531

1.3609

1.7140

250

29.825

60.725

0.01700

13.804

13.821

218.48

945.5

1164.0

0.3675

1.3323

1.6998

260

35.429

72.134

0.01709

11.746

11.763

228.64

938.7

1167.3

0.3817

1.3043

1.6860

270

41.858

85.225

0.01717

10.044

10.061

238.841 931.8

1170.6

0.3958

1.2769

1.6727

Fig. 4-9. Excerpt from typical saturated vapor table.
Reproduced from Thermodynamic Properties of Steam by Keenan and Keyes, published by John Wiley and
Sons, 1936, with permission.

specific volume. If the density only is given
and the specific volume is wanted, the specific
volume is determined by dividing the density
into one. Likewise, when the specific volume is
given and the density is wanted, the density is
found by dividing the specific volume into one
(Section 3-4).

Three values for enthalpy h are usually
given in the saturated vapor tables: (1) the
enthalpy of the liquid (h t ), which is the heat
required to raise the temperature of the liquid
from the temperature at the assumed zero point
of enthalpy to the saturation temperature corre-
sponding to the pressure of the liquid; (2) the
enthalpy of vaporization (h fg ), which is the

latent heat of vaporization at the pressure and
temperature indicated; and (3) the enthalpy of
the vapor (h g ), which is the sum of the enthalpy
of the liquid (h f ) and the enthalpy of vapor-
ization (h fa ). For example, the enthalpy of the
liquid (h f ) for water at 212° F under atmospheric
pressure is 180 Btu (1 x 1 x 180), whereas the
enthalpy of the saturated water vapor at 212° F
under atmospheric pressure is 1050 Btu, which
is the sum of the enthalpy of the liquid (180 Btu)
and the enthalpy of vaporization (970 Btu).

Two values of entropy are usually given:
s f , the entropy of the liquid and s g , the entropy
of the vapor, the difference between the two
being the change in entropy during vaporization.

54 PRINCIPLES OF REFRIGERATION

Dichlorodifluoromethane (Refrigerant-12)
Properties of Superheated Vapor

Abs. Pressure 36 lb/in. s

Abs. Pressure 38 lb/in.*

Abs. Pressure 40 lb/in. 2

Abs. Pressure 42 lb/in. 2

Temp.

Gage Pressure 21.3 lb/in. 8

Gage Pressure 23.3 lb/in. 8

Gage Pressure 25.3 lb/in. 8

Gage Pressure 27.3 lb/in. 8

°F

(Sat. Temp. 20.4° F)

(Sat. Temp. 23.2° F)

(Sat. Temp. 25.9° F)

(Sat. Temp. 28.5° F)

(at
sat'n)

{1.113) {80.54) {0.16947)

{1.058) {80.86) {0.16931)

{1.009) {81.16) (0.16914)

(0,963) (81.44) (0.16897)

1.140

81.90

0.17227

1.076

81.82

0.17126

1.019

81.76

0.17030

0.967

81.65

0.16939

1.168

1.196
1.223

83.35

84.81
86.27

0.17518

0.17806
0.18089

1.103

1.129
1.156

83.27

84.72
86.19

0.17418

0.17706
0.17991

1.044

83.20

0.17322

0.991

1.016
1.040

83.10

84.56
86.03

0.17231

1.070

84.65

0.17612

0.17521

1.095

86.11

0.17896

0.17806

1.250

87.74

0.18369

1.182

87.67

0.18272

1.120

87.60

0.18178

1.063

87.51

0.18086

1.278

89.22

0.18647

1.208

89.16

0.18551

1.144

89.09

0.18455

1.087

89.00

0.18365

1.305

90.71

0.18921

1.234

90.66

0.18826

1.169

90.58

0.18731

1.110

90.50

0.18640

100

1.332

92.22

0.19193

1.260

92.17

0.19096

1.194

92.09

0.19004

1.134

92.01

0.18913

110

1.359

93.75

0.19462

1.285

93.69

0.19365

1.218

93.62

0.19272

1.158

93.54

0.19184

120

1.386

95.28

0.19729

1.310

95.22

0.19631

1.242

95.15

0.19538

1.181

95.09

0.19451

130

1.412

96.82

0.19991

1.336

96.76

0.19895

1.267

96.70

0.19803

1.204

96.64

0.19714

140

1.439

98.37

0.20254

1.361

98.32

0.20157

1.291

98.26

0.20066

1.227

98.20

0.19979

ISO

1.465

99.93

0.20512

1.387

99.89

0.20416

1.315

99.83

0.20325

1.250

99.77

0.20237

160

1.492

101.51

0.20770

1.412

101.47

0.20673

1.340

101.42

0.20583

1.274

101.36

0.20496

170

1.518

103.11

0.21024

1.437

103.07

0.20929

1.364

103.02

0.20838

1.297

102.96

0.20751

180

1.545

104.72

0.21278

1.462

104.67

0.21183

1.388

104.63

0.21092

1.320

104.57

0.21005

190

1.571

106.34

0.21528

1.487

106.29

0.21433

1.412

106.25

0.21343

1.343

106.19

0.21256

200

1.597

107.97

0.21778

1.512

107.93

0.21681

1.435

107.88

0.21592

1.365

107.82

0.21505

210

1.623

109.61

0.22024

1.537

109.57

0.21928

1.459

109.52

0.21840

1.388

109.47

0.21754

220

1.650

111.27

0.22270

1.562

111.22

0.22176

1.482

111.17

0.22085

1.411

111.12

0.22000

230

1.676

112.94

0.22513

1.587

112.89

0.22419

1.506

112.84

0.22329

1.434

112.80

0.22244

240

1.702

114.62

0.22756

1.612

114.58

0.22662

1.530

114.52

0.22572

1.457

114.49

0.22486

250

1.728

116.31

0.22996

1.637

116.28

0.22903

1.554

116.21

0.22813

1.480

116.19

0.22728

260

1.754

118.02

0.23235

1.662

117.99

0.23142

1.577

117.92

0.23052

1.502

117.90

0.22967

270

1.780

119.74

0.23472

1.687

119.71

0.23379

1.601

119.65

0.23289

1.524

119.62

0.23204

280

1.807

121.47

0.23708

1.712

121.45

0.23616

1.625

121.40

0.23526

1.547

121.36

0.23441

290

1.833

123.22

0.23942

1.737

123.20

0.23850

1.649

123.15

0.23760

1.570

123.11

0.23675

300

1.762

124.95

0.24083

1.673

124.92

0.23994

1.592

124.87

0.23909

Fig. 4-10. Excerpt from typical superheated vapor table.
Copyright by E. I. du Pont de Nemours and Co., Inc. Reprinted by permission.

It has been stated previously that the con-
dition of a gas or a vapor can be determined
when any two of its properties are known. How-
ever, for a saturated liquid or vapor at any one
pressure, there is only one temperature that the
fluid can have and still satisfy the conditions of
saturation. This is true also for the other
properties of a saturated liquid or vapor.
Therefore, if any one property of a saturated
liquid or vapor is known, the value of the other
properties can be read directly from the satu-
rated vapor table. For instance, assume that
the pressure of one pound of dry saturated
steam is 20 psia. By locating 20 psia (encircled)
in the second column of the abbreviated table in

Fig. 4-9 and reading across the table, the values
(set off by the heavy lines) for all the other
properties of the vapor at this condition can be
obtained.

4-22. Superheated Vapor Tables. A super-
heated vapor table deals with the properties of
a superheated vapor rather than those of a
saturated vapor, and the arrangement of a
superheated vapor table is somewhat different
from that of a saturated vapor table. One
common form of the superheated vapor table is
illustrated in Fig. 4-10.

Before examining the superheated vapor
table, it is important to take note of one
significant difference between a saturated and a

SATURATED AND SUPERHEATED VAPORS 55

superheated vapor. Whereas, for a saturated
vapor at any one pressure there is only one
temperature which will satisfy the conditions
of saturation, a superheated vapor may have
any temperature above the saturation tempera-
ture Corresponding to its pressure. The specific
volume, enthalpy, and entropy of a superheated
vapor at any one pressure will vary with the
temperature. This does not mean that the
properties of a superheated vapor are entirely
independent of the pressure of the vapor but
only that the properties of the superheated
vapor at any one pressure will vary with the
temperature. As a matter of fact, superheated
vapor tables are based on the pressure of
the vapor, and before the properties of a super-
heated vapor can be determined from a table, the
pressure of the vapor or one of the properties
of the vapor at saturation must be known.
When one of the properties of the vapor at
saturation is known, the pressure of the vapor
can be found by consulting a saturated vapor
table.

In addition to the properties of the super-
heated vapor at various temperatures above the
saturation temperature corresponding to the
pressure, superheated vapor tables usually list
some or all of the properties of the vapor at the
saturation temperature. For example, in Fig.
4-10, the absolute and gage pressures, along
with the saturation temperature corresponding
to these pressures, are given at the head of
the table. The first readings in the body of the
table (italicized) lists the specific volume, the
enthalpy, and the entropy of the vapor at
saturation. The specific volume, enthalpy, and
entropy of the superheated vapor at various
temperatures above the saturation temperature
make up the body of the table. Notice that the
temperature of the superheated vapor, given in
the extreme left-hand column, is listed in 10° F
increments.

Example 4-1. One pound of superheated

Refrigerant-12 vapor is at a temperature of

50° F and its pressure is 40 psia. From the

abbreviated table in Fig. 4-10, determine:

(a) The temperature, volume, enthalpy, and

entropy of the vapor at saturation
(fc) The volume, enthalpy, and entropy of the

vapor at the superheated condition
(c) The degree of superheat of the vapor in
degrees Fahrenheit

(d) The amount of superheat in the vapor in
Btu

(e) The change in the volume during the
superheating

(/) The change in entropy during the super-
heating

Solution

(a) From the head of

the table, the satura-

tion temperature

corresponding to

40 psia

= 25.9° F

From the body of

the table (first

reading, itali-

cized),thespecific

volume of the

vapor at satura-

tion

= 1.009 cu ft/lb

The enthalpy of the

vapor at satura-

tion

= 81.16 Btu/lb

The entropy of the

vapor at satura-

tion

= 0.16914 Btu/lb/° R

(b) From the body

of the table, the

properties of the

vapor superheated

to 50° F (offset by

heavy lines in Fig.

4-10) the specific

volume

= 1.070 cu ft/lb

The enthalpy

= 84.65 Btu/lb

The entropy

= 0.17612 Btu/lb/ R

temperature

= 50.0° F

The temperature at

saturation

= 25.9° F

The degree of super-

heat of the vapor

in degrees Fahr-

enheit

= 24.1°F

(rf)The enthalpy of

the superheated

vapor

= 84.65 Btu/lb

The enthalpy of the

vapor at satura-

tion

= 81.16 Btu/lb

The amount of

superheat in the

vapor in Btu

= 3.49 Btu/lb

56 PRINCIPLES OF REFRIGERATION

(e) The entropy of the

superheated vapor

The entropy of the
vapor at satura-
tion

The change in
entropy during
the superheating

= 0.17612 Btu/lb/°R

= 0.16914 Btu/lb/° R

= 0.00698 Btu/lb/° R

(/) The volume of the
superheated vapor
The volume of the
vapor at satura-
tion
The change in vol-
ume during the
superheating

= 1.070 cu ft/lb
= 1.009 cu ft/lb
= 0.061 cu ft/lb

Psych rometric
Properties of Air

5-1. Composition of Air. Air is a mecha-
nical mixture of gases and water vapor. Dry
air (air without water vapor) is composed
chiefly of nitrogen (approximately 78% by
volume) and oxygen (approximately 21 %), the
remaining 1 % being made up of carbon dioxide
and minute quantities of other gases, such as
hydrogen, helium, neon, argon, etc. With
regard to these dry air components, the com-
position of the air is practically the same
everywhere. On the other hand, the amount of
water vapor in the air varies greatly with the
particular locality and with the weather con-
ditions. Since the water vapor in the air results
primarily from the evaporation of water from
the surface of various bodies of water, atmos-
pheric humidity (water vapor content) is
greatest in regions located near large bodies of
water and is less in the more arid regions.

Since all air in the natural state contains a
certain amount of water vapor, no such thing
as "dry air" actually exists. Nevertheless, the
concept of "dry air" is a very useful one in that
it greatly simplifies psychrometric calculations.
Hereafter in this book the term "dry air" is
used to denote air without water vapor, whereas
the term "air" is used to mean the natural
mixture of "dry air" and water vapor.
5-2. Air Quantities. Air quantities may be
stated either in units of volume (cubic feet) or in
units of weight (pounds) so that the need for
converting air quantities from one unit of
measure to the other occurs frequently.

The volume occupied by any given weight of
air depends upon the pressure and temperature
of the air, and varies inversely with the baro-
metric pressure and directly with the absolute
temperature. Air very nearly approaches the
condition of an ideal gas and will follow the gas
laws with sufficient accuracy for all practical
purposes. Therefore, the volume occupied by
any given weight of air at any given pressure
and temperature can be determined by applying
Equation 3-10.

Example 5-1. Determine the volume oc-
cupied by 1 lb of air having a temperature of
70° F at standard sea level pressure (14.7 psia).

Solution. Rearranging y _ M x R x T
and applying Equation P

3-10, 1 x 53.3 x

(70 + 460)
14.7 x 144
= 13.34 cu ft

Example 5-2. Determine the volume of the
air in Example 5-1 if the barometric pressure is
12.6 psia.

Solution. Applying
Equation 3-10,

V =

1 x 53.3 x
(70 + 460)

12.6 x 144
= 15.57 cu ft

Example 5-3. Determine the volume of the
air in Example 5-1 if the temperature of the air
is 100° F.

Solution. Applying
Equation 3-10,

K =

1 x 53.3 x
(100 + 460)

14.7 x 144
= 14.10 cu ft

The relationship between the volume and the
weight of a given quantity of air at any condition
is expressed by the following equations:

V = M x v (5-1)

V
M = -

(5-2)

where M = the weight of air in pounds

V = the volume of M pounds of air in

cubic feet
v = the specific volume of the air in

cubic feet per pound

Example 5-4. Air at a temperature of
95° F is circulated over a cooling coil at the
rate of 2000 cu ft/min (cfm). If the specific

58 PRINCIPLES OF REFRIGERATION

volume of the air is 14.38 cu ft/lb, determine
the weight of air passing over the coil in pounds
per hour.

Solution. Applying M _

Equation 5-2, the weight
of air passing over the
cooling coil

Multiplying by 60 min

2000

14.38
= 139.2 lb/min

M - 139.1 x 60
= 8346 lb/hr

5-3. Standard Air. Because of the difference
in the volume of any given weight of air at
various temperatures and pressures, an air
standard has been established for use in the
rating of air handling equipment so that all
equipment is rated at equal conditions. Dry air
having a specific volume of 1 3.34 cu ft per pound
or a density of 0.07496 (0.075) lb per cu ft
(1/13.34) is defined as standard air. Air at a
temperature of 70° F and at standard sea level
pressure has this specific volume and density
(see Example 5-1).

A given volume of air at any condition can be
converted to an equivalent volume of standard
air by applying the following equation :

(5-3)

where V s = the equivalent volume of standard
air
V a = the actual volume of the air at any

given condition
v a =* the specific volume of the air at the

given condition
v, = the specific volume of standard air
(13.34 cu ft/lb)

Example 5-5. For the air in Example 5-4,
determine the equivalent volume of standard
air.

Solution. Applying
Equation 5-3, the equiv-
alent volume of standard
airK,

_ 2000 x 14.38

13.34
= 2155 cfm

5-4. Dalton's Law of Partial Pressure.

Dalton's law of partial pressures states in effect
that in any mechanical mixture of gases and
vapors (those which do not combine chemically) :
(1) each gas or vapor in the mixture exerts an
individual partial pressure which is equal to the
pressure that the gas would exert if it occupied
the space alone and (2) the total pressure of the
gaseous mixture is equal to the sum of the

partial pressures exerted by the individual
gases or vapors.

Air, being a mechanical mixture of gases and
water vapor, obeys Dalton's law. Therefore,
the total barometric pressure is always equal to
the sum of the partial pressures of the dry gases
and the partial pressure of the water vapor.
Since psychrometry is the study of the properties
of air as affected by the water vapor content,
the individual partial pressures exerted by the
dry gases are unimportant and, for all practical
-purposes, the total barometric pressure may be
considered to be the sum of only two pressures:
(1) the partial pressure exerted by the dry gases
and (2) the partial pressure exerted by the water
vapor.

5-5. Dew Point Temperature. It is im-
portant to recognize that the water vapor in
the air is actually steam at low pressure and that
this low pressure steam, like high pressure steam
will be in a saturated condition when its
temperature is the saturation temperature corre-
sponding to its pressure. Since all of the com-
ponents in a gaseous mixture are at the same
temperature, it follows that when air is at any
temperature above the saturation temperature
corresponding to the partial pressure exerted by
the water vapor the water vapor in the air will
be superheated. On the other hand, when air
is at a temperature equal to the saturation
temperature corresponding to the partial
pressure of the water vapor, the water vapor in
the air is saturated and the air is said to be
saturated (actually it is only the water vapor
which is saturated). The temperature at which
the water vapor in the air is saturated is known
as the dew point temperature of the air. Ob-
viously, then, the dew point temperature of the
air is always the saturation temperature corre-
sponding to the partial pressure exerted by the
water vapor. Hence, when the partial pressure
exerted by the water vapor is known, the dew
point temperature of the air can be determined
from the steam tables. Likewise, when the dew
point temperature of the air is known, the
partial pressure exerted by the water vapor can
be determined from the steam tables.

Example 5-6. Assume that a certain quan-
tity of air has a temperature of 80° F and that
the partial pressure exerted by the water vapor
in the air is 0.17811 psia. Determine the dew
point temperature of the air.

PSYCHROMETRIC PROPERTIES OF AIR 59

Solution. From Table 4-1, the saturation
temperature of steam corresponding to a pres-
sure of 0.17811 psia is 50° F. Therefore, 50° F
is the dew point temperature of the air.

Example 5-7. A certain quantity of air has
a temperature of 80° F and a dew point tem-
perature of 40° F. Determine the partial pres-
sure exerted by the water vapor in the air.

Solution. From Table 4-1, the saturation
pressure corresponding to 40° F is 0.12170 psia
and therefore 0.12170 psia is the partial pressure
exerted by the water vapor.

It has been shown (Section 4-5) that the
pressure exerted by any vapor is directly
proportional to the density (weight per unit
volume) of the vapor. Since the dew point
temperature of the air depends only on the
partial pressure exerted by the water vapor, it
follows that, for any given volume of air, the
dew point temperature of the air depends only
upon the weight of water vapor in the air. As
long as the weight of water vapor in the air
remains unchanged, the dew point temperature
of the air will also remain unchanged. If the
amount of water vapor in the air is increased or
decreased, the dew point temperature of the air
will also be increased or decreased, respectively.
Increasing the amount of water vapor in the air
will increase the pressure exerted by the water
vapor and raise the dew point temperature.
Likewise, reducing the amount of water vapor
in the air will reduce the pressure of the water
vapor and lower the dew point temperature.
5-6. Maximum Water Vapor Content.
The maximum amount of water vapor that can
be contained in any given volume of air depends
only upon the temperature of the air. Since the
amount of water vapor in the air determines the
partial pressure exerted by the water vapor, it
is evident that the air will contain the maximum
amount of water vapor when the water vapor
in the air exerts the maximum possible pressure.
Since the maximum pressure that can be
exerted by any vapor is the saturation pressure
corresponding to its temperature, the air will
contain the maximum weight of water vapor
when the pressure exerted by the water vapor
is equal to the saturation pressure corresponding
to the temperature of the air. At this condition
the temperature of the air and the dew point
temperature will be one and the same and the air

will be saturated. It is important to notice that
the higher the temperature of the air, the higher
is the maximum possible vapor pressure and the
greater is the maximum possible water vapor
content.

5-7. Absolute Humidity. The water vapor
in the air is called humidity. The absolute
humidity of the air at any given condition is
denned as the actual weight of water vapor
contained in 1 cu ft of air at that condition.
Since the weight of water vapor contained in the
air is relatively small, it is often measured in
grains rather than in pounds (7000 grains equal
lib).

5-8. The Psychrometric Tables. It was
shown in Section 5-5 that the actual weight of
water vapor contained in a unit volume of air is
solely a function of the dew point temperature
of the air. Because of this fixed relationship
between the dew point temperature and the
absolute humidity of the air, when the value
of one is known, the value of the other can be
readily computed.

The absolute humidity of air at various dew
point temperatures is listed in Tables 5-1 and
5-2. The dew point temperatures are listed in
column (1) of the tables, and the absolute
humidity corresponding to each of the dew
point temperatures is given in columns (4) and
(5). The values given in column (4) are in
pounds of water vapor per cubic foot of air,
whereas the values given in column (5) are in
grains of water vapor per cubic foot of air.
Too, the partial pressure (saturation pressure)
of the vapor corresponding to each dew point
temperature is given in inches of mercury in
column (2) and in pounds per square inch in
column (3).

5-9. Relative Humidity. Relative humidity
(RH), expressed in percent, is the ratio of the
actual weight of water vapor per cubic foot of
air relative to the weight of water vapor con-
tained in a cubic foot of saturated air at the
same temperature, viz:

Actual weight of
water vapor per

cubic foot of air

Relative humidity = ... . . . -z — x 100

J Weight of water

vapor in 1 cu ft of

saturated air at the

same temperature

60 PRINCIPLES OF REFRIGERATION

For instance, if air at a certain temperature
contains only half as much water vapor per
cubic foot of air as the air could contain at that
temperature if it were saturated, the relative
humidity of the air is 50%. The relative
humidity of saturated air, of course, is 100%.

Example 5-8. Air at a temperature of
80° F has a dew point temperature of 50° F.
Determine the relative humidity.

Solution. From
Table 5-2, absolute
humidity correspond-
ing to dew point tem-
perature of 50° F

Absolute humidity
of saturated air at
80° F

Applying Equation _
5-4, the relative humid- — 11,04
ity of the air = 37.1 %

= 4.106 grains/cu ft

11.04 grains/cu ft

4.106 1/MV
x 100

Example 5-9. Determine the relative hu-
midity of the air in Example 5-8, if the air is
cooled to 60° F. (Note : the dew point tempera-
ture of the air does not change because the
moisture content does not change.)

Solution. From Table
5-2, absolute humidity
corresponding to dew
point of 50° F

Absolute humidity of
saturated air at 60° F

Applying Equation 5-4,
the relative humidity of
the air

= 4.106 grains/cu ft

= 5.795 grains/cu ft
_ 4.106
5.795
= 70.8%

5-10. Specific Humidity. The specific humid-
ity is the actual weight of water vapor mixed
with 1 lb of dry air and is usually stated in
grains per pound, that is, grains of water vapor
per pound of dry air. For any given barometric
pressure, the specific humidity is a function of
the dew point temperature alone. The specific
humidity of air at various dew point tempera-
tuies is listed in Columns 6 and 7 of Tables 5-1
and 5-2. In Column 6, the specific humidity is
given in pounds of water vapor per pound of dry
air, whereas in Column 7 the specific humidity
is given in grains of water vapor per pound of
dry air. Since the specific humidity correspond-
ing to any given dew point temperature varies
with the total barometric pressure, the values

given in Tables 5-1 and 5-2 apply only to air at
standard barometric pressure.

The specific humidity of the air at any given
dew point temperature increases as the total
barometric pressure decreases and decreases as
the total barometric pressure increases. The
reason for this is easily explained. It has been
shown (Examples 5-1 and 5-3) that the volume
occupied by 1 lb of air increases as the total
barometric pressure decreases. Since the density
of a vapor varies inversely with the volume, it
follows that the weight of water vapor required
to produce a given vapor density and vapor
pressure increases as the volume of the air
increases. Likewise, as the volume occupied by
1 lb of air diminishes, the weight of water vapor
required to produce a certain vapor density and
vapor pressure also diminishes.
5-11. Percentage Humidity. Percentage
humidity is defined as the ratio of the actual
weight of water vapor in the air per pound of dry
air to the weight of water vapor required to
saturate completely 1 lb of dry air at the same
temperature. Percentage humidity, like relative
humidity, is given in percent. Notice, however,
that percentage humidity is associated with the
weight of water vapor per unit weight of air,
whereas relative humidity is associated with the
weight of water vapor per unit volume of air.
For this reason the percentage humidity varies
with the total barometric pressure, whereas
relative humidity does not.

Example 5-10. Air at standard sea level
pressure has a temperature of 80° F and a dew
point temperature of 50° F. Determine the
specific humidity and percentage humidity of
the air.

Solution. From Table
5-2, the specific humidity
of the air in grains per
pound corresponding to
a 50° F dew point tem-
perature (Column 7)

Specific humidity of
saturated air at 80° F
(Column 7)

Percentage humidity

53.38 grains/lb

155.50 grains/lb
53.38

155.50

34.3%

x 100

Note. Compare this value with the relative
humidity obtained in Example 5-8.

PSYCHROMETRIC PROPERTIES OF AIR 61

5-12. Dry Bulb and Wet Bulb Tempera-
tures. The dry bulb (DB) temperature of the
air is the temperature as measured by an ordinary
dry bulb thermometer. When measuring the dry
bulb temperature of the air, the bulb of the ther-
mometer should be shaded to reduce the effects
of direct radiation.

The wet bulb (WB) temperature of the air is
the temperature as measured by a wet bulb ther-
mometer. A wet bulb thermometer is an ordi-
nary thermometer whose bulb is enclosed in a
wetted cloth sac or wick. To obtain an accurate
reading with a wet bulb thermometer, the wick
should be wetted with clean water at approxi-
mately the dry bulb temperature of the air and
the air velocity around the wick should be main-
tained between 1000 and 2000 ft per minute. As
a practical matter, this velocity can be simulated
in still air by whirling the thermometer about on
the end of a chain. An instrument especially
designed for this purpose is the sling psychrom-
eter (Fig. 5-1). The sling psychrometer is made
up of two thermometers, one dry bulb and one
wet bulb, mounted side by side in a protective
case which is attached to a handle by a swivel
connection so that the case can be easily rotated
about the hand. After saturating the wick with
clean water, the instrument is whirled rapidly in
the air for approximately one minute, after which
time readings can be taken from both the dry
bulb and wet bulb thermometers. The process
should be repeated several times to assure that
the lowest possible wet bulb temperature has
been recorded.

Unless the air is 100 % saturated, in which case
the dry bulb, wet bulb, and dew point tempera-
tures of the air will be one and the same, the
temperature recorded by a wet bulb thermom-
eter will always be lower than the dry bulb
temperature of the air. The amount by which
the wet bulb temperature is reduced below the
dry bulb temperature depends upon the relative
humidity of the air and is called the wet bulb
depression.

Whereas a dry bulb thermometer, being un-
affected by humidity, measures only the actual
temperature of the air, a wet bulb thermometer,
because of its wetted wick, is greatly influenced
by the moisture in the air; thus a wet bulb tem-
perature is in effect a measure of the relationship
between the dry bulb temperature of the air and
the moisture content of the air. In general, for

Swivel
connection

Wet bulb
thermometer

.Dry bulb
"thermometer

Wetted wick

Fig. 5-1. Sling psychrometer.

any given dry bulb temperature, the lower the
moisture content of the air, the lower is the wet
bulb temperature. The reason for this is easily
explained.

When unsaturated air is brought into contact
with water, water will evaporate into the air at a
rate proportional to the difference in pressure
between the vapor pressure of the water and the
vapor pressure of the air. Hence, when a wet
bulb thermometer is whirled rapidly about in
unsaturated air, water will evaporate from the
wick, thereby cooling the water remaining in the
wick (and the thermometer bulb) to some tem-
perature below the dry bulb temperature of the
air.

It is important to recognize the fact that the
wet bulb temperature of the air is a measure of
the relationship between the dry bulb and dew
point temperatures of the air, and as such it
provides a convenient means of determining the
dew point temperature of the air when the dry

62 PRINCIPLES OF REFRIGERATION

bulb temperature is known. Too, it will be
shown later that the wet bulb temperature is also
an index of the total heat content of the air.

In order to understand why the wet bulb tem-
perature is a measure of the relationship between
the dry bulb and dew point temperatures, a know-
ledge of the theory of the wet bulb thermometer
is required. When water evaporates from the
wick of a wet bulb thermometer, heat must be
supplied to furnish the latent heat of vaporiza-
tion. Before the temperature of the water in the
wick is reduced below the dry bulb temperature
of the air, the source of the heat to vaporize the
water is the water itself. Therefore, as water
evaporates from the wick, the water remaining
in the wick is cooled below the dry bulb tempera-
ture of the air. When this occurs, a temperature
differential is established and heat begins to flow
from the air to the wick. Under this condition, a
part of the vaporization heat is being supplied by
the air while the other part is supplied by the
water in the wick. As the temperature of the
wick continues to decrease, the temperature
difference between the air and the wick increases
progressively so that more and more of the
vaporization heat is supplied by the air and less
and less is supplied by the water in the wick.
When the temperature of the wick is reduced to
the point where the temperature difference
between the air and the "wick is such that the
flow of heat from the air is sufficient to supply all
of the vaporizing heat, the temperature of the
wick will stabilize even though vaporization from
the wick continues. The temperature at which
the wick stabilizes is called the temperature of
adiabatic saturation and is the wet bulb tempera-
ture of the air.

Through careful analysis of the foregoing, it
can be seen that the wet bulb temperature
depends upon both the dry bulb temperature and
the amount of water vapor in the air. For
example, the lower the relative humidity of the
air, the greater is the rate of evaporation from
the wick and the greater is the amount of heat
required for vaporization. Obviously, the greater
the need for heat, the greater is the wet bulb
depression below the dry bulb temperature. Too,
it follows also that the lower the dry bulb tem-
perature, the lower the wet bulb temperature for
any given wet bulb depression.
5-13. The Heat Content or Enthalpy of

Air. Air has both sensible and latent heat,

and the total heat content of the air at any
condition is the sum of the sensible and latent
heat contained therein.

The sensible heat of the air is a function of the
dry bulb temperature. For any given dry bulb
temperature, the sensible heat of the air is taken
as the enthalpy of dry air at that temperature as
calculated from 0° F. Air sensible heat at various
temperatures is given in Btu per pound of dry air
in Column 10 of Tables 5-1 and 5-2. With regard
to Column 10, the temperatures listed in Column
1 are used as dry bulb temperatures.

Example 5-1 1. Using Table 5-2, determine
the sensible heat in 10 lb of air at 80° F.

Solution. From Table
5-2, the sensible heat of 1 lb
of air at 80° F

For 10 lb of air, the sen-
sible heat at 80° F

= 19.19 Btu/lb
= 10 x 19.19
= 191 .9 Btu

The quantity of sensible heat added or
removed in heating or cooling a given weight of
air through a given temperature range may be
computed by applying Equation 2-8. The mean
specific heat of air at constant pressure is 0.24
Btu/lb. (Although the specific heat of any vapor
or gas varies somewhat with the temperature
range, the use of a mean specific heat value is
sufficiently accurate for all practical purposes.)

Example 5-12. Compute the quantity of
sensible heat required to raise the temperature
of 10 lb of air from 0° F to 80° F.

Solution. Applying
Equation 2-8, Q s

Alternate Solution. From
Table 5-2, the sensible heat
of 1 lb of air at 80° F

Sensible heat of 1 lb of
air at 0° F

For 1 lb of air, Q a

For 10 lb of air, Q,

10 x 0.24

x (80 - 0)
192 Btu

= 19.19 Btu/lb

= Btu/lb
= 19.19 -0
= 19.19 Btu/lb
= 10 x 19.19
= 191.9 Btu

Since all the components of dry air are non-
condensable at normal temperatures and pres-
sures, for all practical purposes the only latent
heat in the air is the latent heat of the water
vapor in the air. Therefore, the amount of

PSYCHROMETRIC PROPERTIES OF AIR 63

latent heat in any given quantity of air depends
upon the weight of water vapor in the air and
upon the latent heat of vaporization of water
corresponding to the saturation temperature of
the water vapor.

Since the saturation temperature of the water
vapor is the dew point temperature of the air,
the dew point temperature determines not only
the weight of water vapor in the air but also the
value of the latent heat of vaporization. Hence,
the latent heat content of the air is a function of
the dew point temperature alone. As long as the
dew point temperature of the air remains un-
changed, the latent heat content of the air also
remains unchanged.

The total heat content of water vapor at
various temperatures as computed from 32° F is
given in Btu per pound in Column 1 1 of Tables
5-1 and 5-2. Although the values given in
Column 11 include the sensible heat of the
liquid above 32° F as well as the latent heat of
vaporization at the given temperature, common
practice is to treat the entire heat content of the
water vapor as latent heat.*

The latent heat content of any given quantity
of air can be computed by multiplying the actual
weight of water vapor in the air in pounds by the
total heat of the water vapor as given in Column
11 of Tables 5-1 and 5-2.

Example 5-13. Compute the latent heat
content of the air in Example 5-12, if the dew
point temperature of the air is 50° F.

Solution. From
Table 5-2, the actual
weight of water vapor
per pound of dry air
(specific humidity) at
50° F DP (Column 6) = 0.007626 lb

Total heat per
pound of saturated
water vapor at 50° F
(Column 11) = 1081.7 Btu/lb

* Although the total heat of the air at any con-
dition is the sum of the sensible and latent heat
contained therein, as a practical matter it is more
convenient to consider the total enthalpy of the air
as being the sum of the enthalpy of the dry air and
the enthalpy of the water vapor mixed with the dry
air. Since the amount of sensible heat is com-
paratively small, the error which accrues from
assuming all of the heat of the vapor to be latent
heat is of no practical consequence.

= 0.007626 x 1081.7
= 8.25 Btu/lb
= 10 x 8.25
= 82.5 Btu

Latent heat per
pound of dry air at
50° F DP

For 10 lb of dry air,
total latent heat

Since the total heat of the air is the sum of the
sensible and latent heat contained therein, the
total heat of the air in Examples 5-12 and 5-13
is the sum of the sensible heat of the dry air, as
computed in Example 5-12, and the latent heat
of the water vapor mixed with the dry air, as
computed in Example 5-13, viz:

Sensible heat of 10 lb of dry
air at 70° F, from Example
5-12

Latent heat of water vapor
mixed with 10 lb of dry air at
50° F DP, from Example 5-13

Total heat of 10 lb of air at
70° F DB and 50° F DP*

- 191.9 Btu

82.5 Btu
191.9 + 82.5
274.4 Btu

5-14. Wet Bulb Temperature as a Measure
of Total Heat. It has been shown in pre-
ceding sections that the sensible heat of the air
(the heat content of the dry air) is a function of
the dry bulb temperature and that the latent heat
of the air (the heat content of the water vapor
mixed with the dry air) is a function of the dew
point temperature. Since, for any given com-
bination of dry bulb and dew point temperatures,
the wet bulb temperature of the air can have only
one value, it is evident that the wet bulb tempera-
ture is an index of the total heat content of the
air. However, it is important to recognize that
although there is only one wet bulb temperature
that will satisfy any given combination of dry
bulb and dew point temperatures, there are
many combinations of dry bulb and dew point
temperatures which will have the same wet bulb
temperature (see Fig. 5-2). This means in effect
that different samples of air having the same wet

* The actual weight of air involved is slightly in
excess of 10 lb, being 10 lb of dry air plus the weight
of water vapor (0.007626 lb) mixed with the dry air.
Too, since the temperature of the water vapor is the
same as that of the dry air (70° F), the water vapor
contains a certain amount of superheat (50° F to
70° F) which is not included in the total heat. How-
ever, since both of these values are very small, the
error incurred by neglecting them has no practical
significance.

64 PRINCIPLES OF REFRIGERATION

Temperature,

°F

Heat Content, Btu/lb

Dry

Dew

Wet

Bulb

Point

Bulb

Sensible

Latent

Total

14.39

11.98

26.37

15.59

10.78

26.37

53.5

16.79

9.58

26.37

17.99

8.38

26.37

45.5

19.19

7.18

26.37

40.5

20.39

5.98

26.37

34.5

21.59

4.78

26.37

Fig. 5-2

bulb temperature have the same total heat, even
though the ratio of sensible to latent heat may be
different for the different samples.

The values of total heat listed for various
temperatures in Column 12 of Tables 5-1 and
5-2 are for 1 lb of saturated air at the tempera-
ture shown. However, since all samples of air,
saturated or unsaturated, having the same wet
bulb temperature have the same total heat, the
values given in Column 12 will apply to any
sample of air when the temperatures listed in
Column 1 are used as wet bulb temperatures.

Example 5-14. If 100 lb of air having an
initial wet bulb temperature of 78° F are cooled
to a final wet bulb temperature of 60° F, deter-
mine the total heat removed from the air during
the cooling process.

Solution. From Table
5-2, total heat of 1 lb of
air corresponding to 78" F
WB (Column 12) - 63.05 Btu/lb

Total heat per pound of
air at 60° F WB (Column
12)

Total heat removed per
pound of air, Q t

For 1001b of air, the
total heat removed, Q t

= 26.37 Btu/lb
= 63.05 - 26.37
= 36.68 Btu/lb

- 100 x 36.68
= 366.8 Btu

5-15. Specific Volume of Air. It has already
been shown that the volume occupied by a given
weight of air depends upon the temperature of
the air and upon the total barometric pressure.
For standard sea level pressure, the volume of
1 lb of dry air at various temperatures is listed
in Column 8 of Tables 5-1 and 5-2. The volume
of 1 lb of saturated air (1 lb of dry air and the
water vapor to saturate it) is listed for various
temperatures in Column 9. When the relative

humidity of the air is known, the specific volume
of partially saturated air at any condition can be
computed by applying these values in the follow-
ing equation:

v a = v* + K»» - v d ) x %RH] (5-5)

where v a = the specific volume of partially

saturated air
v d = the specific volume of dry air at the

same temperature
v s = the specific volume of saturated air

at the same temperature

Example 5-15. Compute the specific vol-
ume of air at 95° F DB and 50% RH.

Solution. From Table
5-2, specific volume of
dry air at 95° F (Column
8)

Specific volume of
saturated air at 95° F
(Column 9)

Applying Equation
5-5, v a

= 1 3.97 cu ft/lb

= 14.79 cu ft/lb
= 13.97 + [(14.79
- 13.97) x 0.5]
= 14.38 cu ft/lb

5-16. The Psychrometric Chart. Psychro-
metric charts (Fig. 5-3) are graphical repre-
sentations of psychrometric data such as those
contained in Tables 5-1 and 5-2. The use of
psychrometric charts permits graphical analysis
of psychrometric data and thereby facilitates the
solution of many practical problems dealing with
air which would otherwise require tedious mathe-
matical calculation.

Basically, the psychrometric chart shows the
relationship between four fundamental proper-
ties of air: (1) dry bulb temperature, (2) dew
point temperature, (3) wet bulb temperature, and
(4) relative humidity. When any two of these
four properties are known, the other two can be
determined directly from the psychrometric chart
without using mathematical calculations.

The skeleton chart in Fig. 5-4 illustrates the
general construction of the psychrometric chart
which is based primarily upon the relationship
that exists between the aforementioned four pro-
perties. Notice that the lines of dry bulb tem-
perature are vertical while the lines of dew point
temperature are horizontal. The lines of wet
bulb temperature run diagonally across the chart
as do the lines of constant volume. The curved

PSYCHROMETRIC PROPERTIES OF AIR 65

lines are lines of constant relative humidity. The
curved line bounding the chart on the left side is
the line of 100% relative humidity and is called
the saturation curve. Air at any condition such
that its state can be identified by a point falling
anywhere along the saturation curve is saturated
air. Values for dry bulb, wet bulb, and dew
point temperatures are read at the saturation
curve. Values for dry bulb temperature are also
given at the base of the chart. Notice that the
dry bulb, wet bulb, and dew point temperatures
for saturated air coincide. Values of specific
volume and relative humidity are given along
the lines of constant volume and relative humid-
ity, respectively. Values of specific humidity and
vapor pressure are given on the right and left
margins of the chart. For any given air condi-
tion, the specific humidity and vapor pressure
corresponding to the dew point temperature can
be determined by following the dew point tem-
perature line to the specific humidity and vapor
pressure scales. The total heat corresponding to
any wet bulb temperature is found by following
the wet bulb lines to the total heat scale above
the saturation curve. The following example will
illustrate the use of the psychrometric chart.

Example 5-16. A certain quantity of air has
a dry bulb temperature of 95° F and a wet bulb
temperature of 77° F. From the psychrometric
chart determine all of the following values: (1)
dew point temperature, (2) specific humidity,
(3) vapor pressure, (4) specific volume, (5) total
heat, and (6) relative humidity.

Solution. Using the two known properties of
the air as coordinates the condition of the air
can be established as a point on the chart. Once
this point has been established, the other pro-
perties of the air at this condition can be read
directly from the chart as shown in Fig. 5-5, viz:

Dew point temperature = 70° F

Specific humidity =110 grains/lb

Vapor pressure = 0.37 psia

Specific volume = 14.33 cu ft/lb

Total heat = 40.5 Btu/lb

Relative humidity = 45 %

Example 5-17. For the air in Example 5-16,
determine: (a) the sensible heat per pound of
air and (6) the latent heat per pound of air.

Solution,
(a) From Table 5-2, en-
thalpy of 1 lb of dry air
at 95° F DB, Q, = 22.80 Btu/lb

(b) From Example 5-16,
total heat per pound of
air, Q t

The latent heat per
pound of air, &

= 40.50 Btu/lb
= Qt - Q,

= 40.50 - 22.80
= 17.7 Btu/lb

Example 5-18. If the air in Example 5-16 is
cooled to 75° F, determine:

(a) The final dew point temperature

(b) The final wet bulb temperature

(d) The final total heat per pound

Solution. Since the air is not cooled below the
initial dew point temperature, no moisture is
removed from the air. Therefore, the specific
humidity, dew point temperature, and latent
heat of the air remain unchanged. Hence, the
initial dew point temperature and the new dry
bulb temperature can be used as coordinates to
locate the new condition of the air on the
psychrometric chart (point B in Fig. 5-6). The
following properties of the air at the new con-
dition are taken from the psychrometric chart
as indicated in Fig. 5-6:

(a) Wet bulb temperature = 71.4° F

(b) Relative humidity = 85 %

Example 5-19. With respect to Fig. 5-6, in
cooling the air from condition "A," as described
in Example 5-16, to condition "B," as described
in Example 5-18, compute:

(a) The total heat removed per pound of air

(b) The sensible heat removed per pound of
air.

Solution.
(a) From Example 5-16,
air total heat at A

From Example 5-18, air

total heat at B

The total heat removed

per pound of air

in cooling from A

toB

= 40.50 Btu/lb
= 35.69 Btu/lb

= 40.50 - 35.69
= 4.81 Btu/lb

(b) Since there is no change in the latent heat of
the air, the sensible heat removed per pound
of air is equal to the total heat removed
per pound of air.

Example 5-20. Assume that the air in
Example 5-16 is cooled to 40° F and determine:

(a) The total heat removed per pound

(b) The sensible heat removed per pound

66 PRINCIPLES OF REFRIGERATION

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68 PRINCIPLES OF REFRIGERATION

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70 PRINCIPLES OF REFRIGERATION

Solution. Since the air is cooled below the
dew point temperature, moisture will be con-
densed out of the air and the air at the final
condition will be saturated. Therefore, the dry
bulb temperature, the dew point temperature,
and the wet bulb temperature will coincide at
40° F and the relative humidity of the air will be
100%. On the psychrometric chart, the con-
dition of the air falls on the saturation curve at
40° F (point B in Fig. 5-7).

(a) From the psychromet-
ric chart, the total heat
of the air at the initial

condition (point A) = 40.50 Btu/lb

The total heat of the air

at the final condition

(point B) =15.19 Btu/lb

The total heat removed 40.50 - 15.19

per pound = 25.31 Btu/lb

(b) Applying Equation 2-8,

the sensible heat re- = 1 x 0.24
moved per pound of x (95 — 40)

air =13.2 Btu/lb

PROBLEMS

1. Determine the volume occupied by 1 lb of air
having a temperature of 80° F at standard sea
level pressure. Ans. 13.59 cu ft

2. Compute the volume of the air in Problem 1
if the barometric pressure is 13.5 psia.

Ans. 14.80 cu ft

3. Determine the volume of the air in Problem 1
if the temperature of the air is 120° F.

Ans. 14.60 cu ft

4. Air at a temperature of 90° F is circulated
over a cooling coil at the rate of 1000 cu ft per

min (cfm). If the specific volume of the air is
14.10 cu ft/lb, compute the weight of air passing
over the coil in pounds per hour.

Ans. 4255 lb/hr

5. Compute the equivalent volume of standard
air for the conditions of Problem 4.

Ans. 946 cfm

6. Compute the quantity of sensible heat re-
quired to raise the temperature of 10 lb of air
from a temperature of 35° F to a temperature of
100° F. Ans. 156 Btu

7. If 80 lb of air having an initial wet bulb tem-
perature of 80° F are cooled to a final wet bulb
temperature of 65° F, determine the total heat
removed from the air during the cooling process.

Ans. 1085.6 Btu

8. A certain quantity of air has "a dry bulb
temperature of 90° F and a wet bulb tempera-
ture of 77° F. From the psychrometric chart
determine all of the following values:

(a) dew point temperature, (£>) specific humidity,
(c) vapor pressure, (d) specific volume, (e) total
heat, and (/) relative humidity.
Ans. (a) 72.3° F; (6) 119.5 gpp; (c) 0.395 psia;
(a") 14.23 cu ft/lb; (e) 40.5 Btu/lb; (/) 58%

9. Assume that the air in Problem 8 is cooled to
75° F and determine:

(a) the final dew point temperature Ans. 72.3° F

(b) the final wet bulb temperature Ans. 73° F

Ans. 36.6 Btu/lb

10. Assume that the air in Problem 8 is cooled
to 55° F and determine:

(a) the total heat removed per pound of air

Ans. 17.2 Btu/lb

(6) the sensible heat removed per pound of air

Ans. 8.40 Btu/lb

Ans. 8.8 Btu/lb

Refrigeration
and the Vapor
Compression
System

6-1. Refrigeration. In general, refrigeration
is denned as any process of heat removal. More
specifically, refrigeration is denned as that
branch of science which deals with the process
of reducing and maintaining the temperature of
a space or material below the temperature of the
surroundings.

To accomplish this, heat must be removed
from the body being refrigerated and transferred
to another body whose temperature is below that
of the refrigerated body. Since the heat removed
from the refrigerated body is transferred to
another body, it is evident that refrigerating and
heating are actually opposite ends of the same
process. Often only the desired result distin-
guishes one from the other.
6-2. Need for Thermal Insulation. Since
heat will always travel from a region of high
temperature to a region of lower temperature,
there is always a continuous flow of heat into
the refrigerated region from the warmer sur-
roundings. To limit the flow of heat into the
refrigerated region to some practical minimum,
it is usually necessary to isolate the region from
its surroundings with a good heat insulating
material.

6-3. The Heat Load. The rate at which heat
must be removed from the refrigerated space or

material in order to produce and maintain the
desired temperature conditions is called the heat
load. In most refrigerating applications the total
heat load on the refrigerating equipment is the
sum of the heat that leaks into the refrigerated
space through the insulated walls, the heat that
enters the space through door openings, and the
heat that must be removed from the refrigerated
product in order to reduce the temperature of
the product to the space or storage conditions.
Heat given off by people working in the re-
frigerated space and by motors, lights, and other
electrical equipment also contributes to the load
on the refrigerating equipment.

Methods of calculating the heat load are
discussed in Chapter 10.
6-4. The Refrigerating Agent. In any re-
frigerating process the body employed as the
heat absorber or cooling agent is called the
refrigerant.

All cooling processes may be classified as
either sensible or latent according to the effect
the absorbed heat has upon the refrigerant.
When the absorbed heat causes an increase in
the temperature of the refrigerant, the cooling
process is said to be sensible, whereas when the
absorbed heat causes a change in the physical
state of the refrigerant (either melting or vapor-
izing), the cooling process .is said to be latent.
With either process, if the refrigeratingeffectisto
be continuous, the temperature of the refriger-
ating agent must be maintained continuously
below that of the space or material being
refrigerated.

To illustrate, assume that 1 lb of water at
32° F is placed in an open container inside an
insulated space having an initial temperature of
70° F (Fig. 6-1). For a time, heat will flow from
the 70° F space into the 32° F water and the
temperature of the space will decrease. How-
ever, for each one Btu of heat that the water
absorbs from the space, the temperature of the
water will increase 1° F, so that as the tempera-
ture of the space decreases, the temperature of
the water increases. Soon the temperatures of
the water and the space will be exactly die same
and no heat transfer will take place. Refrigera-
tion will not be continuous because the tempera-
ture of the refrigerant does not remain below the
temperature of the space being refrigerated.

Now assume that 1 lb of ice, also at 32° F, is
substituted for the water (Fig. 6-2). This time

72 PRINCIPLES OF REFRIGERATION

Insulation

Heat leaking
through insulation

Fig. 6-1. Heat flows from warm space to cold water.
Water temperature rises as space temperature
decreases. Refrigeration will not be continuous.

the temperature of the refrigerant does not
change as it absorbs heat from the space. The
ice merely changes from the solid to the liquid
state while its temperature remains constant at
32° F. The heat absorbed by the ice leaves the
space in the water going out the drain and the
refrigerating effect will be continuous until all
the ice has melted.

It is both possible and practical to achieve
continuous refrigeration with a sensible cooling
process provided that the refrigerant is con-
tinuously chilled and recirculated through the
refrigerated space as shown in Fig. 6-3.

Latent cooling may be accomplished with
either solid or liquid refrigerants. The solid
refrigerants most frequently employed are ice
and solid carbon dioxide (dry ice). Ice, of
course, melts into the liquid phase at 32° F,
whereas solid carbon dioxide sublimes directly
into the vapor phase at a temperature of —109° F
under standard atmospheric pressure.
6-5. Ice Refrigeration. Melting ice has been
used successfully for many years as a refrigerant.
Not too many years ago ice was the only cooling
agent available for use in domestic and small
commercial refrigerators.

In a typical ice refrigerator (Fig. 6-4) the heat
entering the refrigerated space from all the
various sources reaches the melting ice primarily
by convection currents set up in the air of the
refrigerated space. The air in contact with the
warm product and walls of the space is heated

by heat conducted to it from these materials.
As the air is warmed it expands and rises to the
top of the space carrying the heat with it to the
ice compartment. In passing over the ice the air
is cooled as heat is conducted from the air to the
ice. On cooling, the air becomes more dense and
falls back into the storage space, whereupon it
absorbs more heat and the cycling continues.
The air in carrying the heat from the warm walls
and stored product to the melting ice acts as a
heat transfer agent.

To insure adequate air circulation within the
refrigerated space, the ice should be located near
the top of the refrigerator and proper baffling
should be installed to provide direct and un-
restricted paths of air flow. A drip pan must be
located beneath the ice to collect the water which
results from the melting.

Ice has certain disadvantages which tend to
limit its usefulness as a refrigerant. For instance,
with ice it is not possible to obtain the low tem-
peratures required in many refrigeration applica-
tions. Ordinarily, 32° F is the minimum tem-
perature obtainable through the melting of ice
alone. In some cases, the melting temperature
of the ice can be lowered to approximately 0° F
by adding sodium chloride or calcium chloride
to produce a freezing mixture.

Some of the other more obvious disadvantages
of ice are the necessity of frequently replenishing
the supply, a practice which is neither convenient
nor economical, and the problem of disposing of
the water resulting from the melting.

Insulation

Heat leaking
through insulation

Fig. 6-2. Heat flows from warm space to cold ice.
Temperature of space decreases as ice melts. Tem-
perature of ice remains at 32° F. Heat absorbed by
ice leaves space in water going out the drain.

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 73

Another less obvious, but more important,
disadvantage of employing ice as a refrigerant is
the difficulty experienced in controlling the rate
of refrigeration, which in turn makes it difficult
to maintain the desired low temperature level
within the refrigerated space. Since the rate at
which the ice absorbs heat is directly propor-
tional to the surface area of the ice and to the

6-6. Liquid Refrigerants. The ability of
liquids to absorb enormous quantities of heat as
they vaporize is the basis of the modern mechani-
cal refrigerating system. As refrigerants, vapor-
izing liquids have a number of advantages over
melting solids in that the vaporizing process is
more easily controlled, that is, the refrigerating
effect can be started and stopped at will, the rate

Fig. 6-3. Continuous sensible cooling. Heat taken in by the water in the space is given up to the ice.

temperature difference between the space tem-
perature and the melting temperature of the ice,
the rate of heat absorption by the ice diminishes
as the surface area of the ice is diminished by the
melting process. Naturally, when the refriger-
ating rate diminishes to the point that the heat
is not being removed at the same rate that it is
accumulating in the space from the various heat
sources, the temperature of the space will
increase.

Despite its disadvantages, ice is preferable to
mechanical refrigeration in some applications.
Fresh vegetables, fish, and poultry are often
packed and shipped in cracked ice to prevent
dehydration and to preserve appearance. Too,
ice has tremendous eye appeal and can be used
to considerable advantage in the displaying and
serving of certain foods such as salads, cocktails,
etc., and in chilling beverages.

of cooling can be predetermined within small
limits, and the vaporizing temperature of the
liquid can be governed by controlling the pres-
sure at which the liquid vaporizes. Moreover,
the vapor can be readily collected and condensed
back into the liquid state so that the same liquid
can be used over and over again to provide a
continuous supply of liquid for vaporization.

Until now, in discussing the various properties
of fluids, water, because of its familiarity, has
been used in all examples. However, because of
its relatively high saturation temperature, and
for other reasons, water is not suitable for use as
a refrigerant in the vapor-compression cycle. In
order to vaporize at temperatures low enough to
satisfy most refrigeration requirements, water
would have to vaporize under very low pres-
sures, which are difficult to produce and main-
tain economically.

74 PRINCIPLES OF REFRIGERATION

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Fig. 6-4. Ice refrigerator. Heat is carried from
warm walls and product to the ice by air circulation
within the refrigerated space. Air circulation is by
gravity.

There are numerous other fluids which have
lower saturation temperatures than water at the
same pressure. However, many of these fluids
have other properties that render them unsuitable
for use as refrigerants. Actually, only a relatively
few fluids have properties that make them desir-
able as refrigerants, and most of these have been
compounded specially for that purpose.

There is no one refrigerant which is best suited
for all the different applications and operating
conditions. For any specific application the
refrigerant selected should be the one whose
properties most closely fit the particular require-
ments of the application.

Of all of the fluids now in use as refrigerants,
the one fluid which most nearly meets all the
qualifications of the ideal general-purpose re-
frigerant is a fluorinated hydrocarbon of the
methane series having the chemical name di-
chlorodifluoromethane (CC1 2 F 2 ). It is one of a
group of refrigerants introduced to the industry
under the trade name of "Freon," but is now
manufactured under several other proprietary
designations. To avoid the confusion inherent
jn the use of proprietory or chemical names, this
compound is now referred to as Refrigerant-12.
Refrigerant-12 (R-12) has a saturation tempera-
ture of — 21. 6° F at standard atmospheric
pressure. For this reason, R-12 can be stored as

a liquid at ordinary temperatures only if confined
under pressure in heavy steel cylinders.

Table 16-3 is a tabulation of the thermo-
dynamic properties of R-12 saturated liquid and
vapor. This table lists, among other things, the
saturation temperature of R-12 corresponding
to various pressures. Tables 16-4 through 16-6
list the thermodynamic properties of some of the
other more commonly used refrigerants. These
tables are similar to the saturated liquid and
vapor tables previously discussed and are em-
ployed in the same manner.
6-7. Vaporizing the Refrigerant. An in-
sulated space can be adequately refrigerated
by merely allowing liquid R-12 to vaporize in a
container vented to the outside as shown in Fig.
6-5. Since the R-12 is under atmospheric pres-
sure, its saturation temperature is — 21. 6° F.
Vaporizing at this low temperature, the R-12
readily absorbs heat from the 40° F space
through the walls of the containing vessel. The
heat absorbed by the vaporizing liquid leaves the
space in the vapor escaping through the open
vent. Since the temperature of the liquid
remains constant during the vaporizing process,
refrigeration will continue until all the liquid is
vaporized.

Any container, such as the one in Fig. 6-5, in

js^v,

Refrigerant vapor
/ at atmospheric
( pressure

Fig. 6-5. The Refrigerant-12 liquid vaporizes as it
takes in heat from the 40° F space. The heat taken in
by the refrigerant leaves the space in the vapor
escaping through the vent.

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 75

which a refrigerant is vaporized during a re-
frigerating process is called an evaporator and
is one of the essential parts of any mechanical
refrigerating system.

6-8. Controlling the Vaporizing Tempera-
ture. The temperature at which the liquid
vaporizes in the evaporator can be controlled by
controlling the pressure of the vapor over the
liquid, which in turn is governed by regulating
the rate at which the vapor escapes from the
evaporator (Section 4-5). For example, if a hand
valve is installed in the vent line and the vent is
partially closed off so that the vapor cannot
escape freely from the evaporator, vapor will
collect over the liquid causing the pressure in the
evaporator to rise with a corresponding increase
in the saturation temperature of the refrigerant
(Fig. 6-6). By carefully adjusting the vent valve
to regulate the flow of vapor from the evaporator,
it is possible to control die pressure of the vapor
over the liquid and cause the R-12 to vaporize at
any desired temperature between —21.6° F and
the space temperature. Should the vent valve be
completely closed so that no vapor is allowed to
escape from the evaporator, the pressure in the
evaporator will increase to a point such that the
saturation temperature of the liquid will be equal
to the space temperature, or 40° F. When this
occurs, there will be no temperature differential

Refrigerant

vapor above

atmospheric

pressure

Refrigerant-12
liquid boiling at 30*F

Pressure of

refrigerant vapor

below atmospheric

Refrigerant-12 liquid
'boiling at -100"F

Fig. 6-6. The boiling temperature of the liquid
refrigerant in the evaporator is controlled by
controlling the pressure of the vapor over the liquid
with the throttling valve in the vent.

Fig. 6-7. Pressure of refrigerant in evaporator
reduced below atmospheric by action of a vapor
pump.

and no heat will flow from the space to the
refrigerant. Vaporization will cease and no
further cooling will take place.

When vaporizing temperatures below —21.6°
F are required, it is necessary to reduce the
pressure in the evaporator to some pressure
below atmospheric. This can be accomplished
through the use of a vapor pump as shown in
Fig. 6-7. By this method, vaporization of the
liquid R-12 can be brought about at very low
temperatures in accordance with the pressure-
temperature relationships given in Table 16-3.
6-9. Maintaining a Constant Amount of
Liquid in the Evaporator. Continuous
vaporization of the liquid in the evaporator
requires that the supply of liquid be continuously
replenished if the amount of liquid in the
evaporator is to be maintained constant. One
method of replenishing the supply of liquid in
the evaporator is through the use of a float valve
assembly as illustrated in Fig. 6-8. The action
of the float assembly is to maintain a constant
level of liquid in the evaporator by allowing
liquid to flow into the evaporator from the
storage tank or cylinder at exactly the same rate
that the supply of liquid in the evaporator is
being depleted by vaporization. Any increase in
the rate of vaporization causes the liquid level in
the evaporator to drop slightly, thereby opening

76 PRINCIPLES OF REFRIGERATION

High pressure
S liquid
refrigerant

Needle valve
~ assembly

_ Low pressure
liquid refrigerant

KaSSsiSfc''

Fig. 6-8. Float valve assembly maintains constant
liquid level in evaporator. The pressure of the
refrigerant is reduced as the refrigerant passes
through the needle valve.

the needle valve wider and allowing liquid to
flow into the evaporator at a higher rate. Like-
wise, any decrease in the rate of vaporization
causes the liquid level to rise slightly, thereby
moving the needle valve in the closing direction
to reduce the flow of liquid into the evaporator.
When vaporization ceases entirely, the rising
liquid level will close the float valve tightly and
stop the flow of liquid completely. When vapori-
zation is resumed, the liquid level will fall allow-
ing the float valve to open and admit liquid to
the evaporator.

The liquid refrigerant does not vaporize in the
storage cylinder and feed line because the pres-
sure in the cylinder is such that the saturation
temperature of the refrigerant is equal to the
temperature of the surroundings (see Section
4-10). The high pressure existing in the cylinder
forces the liquid to flow through the feed line
and the float valve into the lower pressure
evaporator. In passing through the float valve,
the high pressure refrigerant undergoes a pres-
sure drop which reduces its pressure to the
evaporator pressure, thereby permitting the re-
frigerant liquid to vaporize in the evaporator at
the desired low temperature.

Any device, such as the float valve illustrated

in Fig. 6-8, used to regulate the flow of liquid
refrigerant into the evaporator is called a
refrigerant flow control. The refrigerant flow
control is an essential part of every mechanical
refrigerating system.

There are five different types of refrigerant
flow controls, all of which are in use to some
extent at the present time. Each of these
distinct types is discussed at length in Chapter
17. The float type of control illustrated in Fig.
6-8 has some disadvantages, mainly bulkiness,
which tend to limit its use to some few special
applications. The most widely used type of
refrigerant flow control is the thermostatic
expansion valve. A flow diagram illustrating the
use of a thermostatic expansion valve to control
the flow of refrigerant into a serpentine coil
type evaporator is shown in Fig. 6-9.
6-10. Salvaging the Refrigerant. As a matter
of convenience and economy it is not practical
to permit the refrigerant vapor to escape to the
outside and be lost by diffusion into the air.
The vapor must be collected continuously and
condensed back into the liquid state so that the
same refrigerant is used over and over again,
thereby eliminating the need for ever replenish-
ing the supply of refrigerant in the system. To

High pressure
liquid

Low pressure

Hiquid-vapor

mixture

Fig. 6-9. Serpentine coil evaporator with thermo-
static expansion valve refrigerant control.

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 77

provide some means of condensing the vapor,
another piece of equipment, a condenser, must
be added to the system (Fig. 6-10).

Since the refrigerant vaporizes in the evapor-
ator because it absorbs the necessary latent heat
from the refrigerated space, all that is required
in order to condense the vapor back into the
liquid state is that the latent heat be caused to
flow out of the vapor into another body. The
body of material employed to absorb the latent
heat from the vapor, thereby causing the vapor
to condense, is called the condensing medium.
The most common condensing media are air
and water. The water used as a condensing
medium is usually supplied from the city main
or from a cooling tower. The air used as a
condensing medium is ordinary outdoor air at
normal temperatures.

For heat to flow out of the refrigerant vapor
into the condensing medium the temperature
of the condensing medium must be below that
of the refrigerant vapor. However, since the
pressure and temperature of the saturated
vapor leaving the evaporator are the same as
those of the vaporizing liquid, the temperature
of the vapor will always be considerably below
that of any normally available condensing
medium. Therefore, heat will not flow out of
the refrigerant vapor into the air or water used
as the condensing medium until the saturation
temperature of the refrigerant vapor has been
increased by compression to some temperature
above the temperature of the condensing
medium. The vapor pump or compressor
shown in Fig. 6-10 serves this purpose.

Before compression, the refrigerant vapor is
at the vaporizing temperature and pressure.
Since the pressure of the vapor is low, the
corresponding saturation temperature is also
low. During compression the pressure of the
vapor is increased to a point such that the
corresponding saturation temperature is above
the temperature of the condensing medium being
employed. At the same time, since mechani-
cal work is done on the vapor in compressing
it to the higher pressure, the internal energy of
the vapor is increased with a corresponding
increase in the temperature of the vapor.

After compression, the high-pressure, high-
temperature vapor is discharged into the con-
denser where it gives up heat to the lower
temperature condensing medium. Since a vapor

low-pressure,
low-temperature

Low-pressure,

low-temperature

liquid-vapor mixture

Refrigerant control -s /

high-temperature

High-pressure,
high-tempenhii

High-pressure, .
high-temperature'
liquid-vapor mixture

High-pressure,

Mgh-temperitun! -

liquid

Fig. 6-10. Collecting and condensing the refrigerant
vapor. Refrigerant absorbs heat in evaporator and
gives off heat in the condenser.

cannot be cooled to a temperature below its
saturation temperature, the continuous loss of
heat by the refrigerant vapor in the condenser
causes the vapor to condense into the liquid
state at the new, higher pressure and saturation
temperature. The heat given off by the vapor in
the condenser is carried away by the condensing
medium. The resulting condensed liquid, whose
temperature and pressure will be the same as
those of the condensing vapor, flows out of the
condenser into the liquid storage tank and is
then ready to be recirculated to the evaporator.
Notice that the refrigerant, sometimes called
the working fluid, is merely a heat transfer
agent which carries the heat from the refriger-
ated space to the outside. The refrigerant
absorbs heat from the refrigerated space in the
evaporator, carries it out of the space, and
rejects it to the condensing medium in the
condenser.

78 PRINCIPLES OF REFRIGERATION

6-11. Typical Vapor-Compression System.

A flow diagram of a simple vapor-compression
system is shown in Fig. 6-11. The principal
parts of the system are: (1) an evaporator,
whose function it is to provide a heat transfer
surface through which heat can pass from the
refrigerated space or product into the vaporizing
refrigerant; (2) a suction line, which conveys
the low pressure vapor from the evaporator
to the suction inlet of the compressor; (3) a
vapor compressor, whose function it is to
remove the vapor from the evaporator, and to
raise the temperature and pressure of the vapor
to a point such that the vapor can be condensed
with normally available condensing media; (4)
a "hot-gas" or discharge line which delivers the
high-pressure, high-temperature vapor from the
discharge of the compressor to the condenser;
(5) a condenser, whose purpose it is to provide
a heat transfer surface through which heat
passes from the hot refrigerant vapor to the
condensing medium; (6) a receiver tank, which
provides storage for the liquid condenser so that
a constant supply of liquid is available to the
evaporator as needed; (7) a liquid line, which
carries the liquid refrigerant from the receiver
tank to the refrigerant flow control; (8) a
refrigerant flow control, whose function it is to
meter the proper amount of refrigerant to the
evaporator and to reduce the pressure of the
liquid entering the evaporator so that the liquid
will vaporize in the evaporator at the desired low
temperature.

4-12. Service Valves. The suction and dis-
charge sides of the compressor and the outlet
of the receiver tank are usually equipped with
manual shut-off valves for use during service
operations. These valves are known as the
"suction service valve," the discharge service
valve," and the "receiver tank valve," respec-
tively. Receiver tanks on large systems frequent-
ly have shut-off valves on both the inlet and the
outlet.

6-13. Division of the System. A refriger-
ating system is divided into two parts according
to the pressure exerted by the refrigerant in the
two parts. The low pressure part of the system
consists of the refrigerant flow control, the
evaporator, and the suction line. The pressure
exerted by the refrigerant in these parts is the
low pressure under which the refrigerant is
vaporizing in the evaporator. This pressure is
known variously as the "low side pressure,"
the "evaporator pressure," the "suction pres-
sure," or the "back pressure." During service
operations this pressure is usually measured
at the compressor by installing a compound
gage on the gage port of the suction service
valve.

The high pressure side or "high side" of the
system consists of the compressor, the discharge
or "hot gas" line, the condenser, the receiver
tank, and the liquid line. The pressure exerted
by the refrigerant in this part of the system is
the high pressure under which the refrigerant is
condensing in the condenser. This pressure is

Suction
line

ine~V.
®

Refrigerant
flow control"!

Evaporator-]

Suction _
service valve

Compressor

■<8K

"SB", rDischargeline

valve | (g) (-Condenser

Liquid

dj~ ,ine

Receiver
J/tank valve

®
VJteceiver
tank

Fig. 6-11. Flow diagram of simple
vapor compression system show-
ing the principal parts.

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 79

Fig, 6-12. Air-cooled condens-
ing unit. Note fan mounted on
motor shaft to circulate air
over condenser.

Receiver
tank

Compressor

ir- coo led
condenser

Compressor
driver

called the "condensing pressure," the "dis-
charge pressure," or, more often, the "head
pressure."

The dividing points between the high and low
pressure sides of the system are the refrigerant
flow control, where the pressure of the refriger-
ant is reduced from the condensing pressure to
the vaporizing pressure, and the discharge valves
in the compressor, through which the high
pressure vapor is exhausted after compression.*
It should be noted that, although the compressor
is considered to be a part of the high side of the
system, the pressure on the suction side of the
compressor and in the crankcase is the low side
pressure. The change in pressure, of course,
occurs in the cylinder during the compression
process.

6-14, Condensing Units. The compressor,
hot gas line, condenser, and receiver tank,
along with the compressor driver (usually an
electric motor), are often combined into one
compact unit as shown in Fig. 6-12. Such an
assembly is called a condensing unit because its
function in the system is to reclaim the vapor
and condense it back into the liquid state.

Condensing units are often classified accord-

• Care should be taken not to confuse the suction
and discharge valves in the compressor with the
suction and discharge service valves. The suction
and discharge valves in a reciprocating compressor
perform the same function as the intake and exhaust
valves in an automobile engine and are vital to the
operation of the compressor, whereas the suction and
discharge service valves serve no useful purpose inso-
far as the operation of the compressor is concerned.
The latter valves are used only to facilitate service
operations, as their nomenclature implies.

ing to condensing medium used to condense the
refrigerant. A condensing unit employing air as
the condensing medium (Fig. 6-12) is called an
air-cooled condensing unit, whereas one employ-
ing water as the condensing medium is a water-
cooled condensing unit.

6-IS. Hermetic Motor-Compressor Assem-
blies. Condensing units of small horsepower
are often equipped with hermetically sealed
motor-compressor assemblies. The assembly
consists of a direct-driven compressor mounted
on a common shaft with the motor rotor and
the whole assembly hermetically sealed in a
welded steel shell (Fig. 6-13).

Condensing units equipped with hermetically
sealed motor-compressor assemblies are known
as "hermetic condensing units" and are em-
ployed on a number of small commercial
refrigerators and on almost all household
refrigerators, home freezers, and window air
conditioners. For reasons that will be shown
later, many hermetic condensing units are not
equipped with receiver tanks,

A variation of the hermetic motor-compressor
assembly is the "accessible hermetic." It is
similar to the full hermetic except that the shell
enclosing the assembly is bolted together rather
than seam welded. (Fig. 6-14). The bolted
construction permits the assemblies to be
opened in the field for servicing.
6-16. Definition of a Cycle. As the refriger-
ant circulates through the system, it passes
through a number of changes in state or
condition, each of which is called a process.
The refrigerant starts at some initial state or
condition, passes through a series of processes
in a definite sequence, and returns to the initial

80 PRINCIPLES OF REFRIGERATION

Fig. 4-13. Air-cooled condensing unit employing hermetic motor-compressor. Note separate fan to circulate
air over condenser. (Courtesy Tecumseh Products Company.)

condition. This series of processes is called a
cycle. The simple vapor-compression refriger-
ation cycle is made up of four fundamental
processes: (1) expansion, (2) vaporization, (3)
compression, and (4) condensation.

To understand properly the refrigeration
cycle it is necessary to consider each process in
the cycle both separately and in relation to the
complete cycle. Any change in any one process
in the cycle will bring about changes in all the
other processes in the cycle.
6-17. Typical Vapor-Compress ton Cycle.
A typical vapor-compression cycle is shown in
Fig. 6-15. Starting at the receiver tank, high-
temperature, high-pressure liquid refrigerant
flows from the receiver tank through the liquid
line to the refrigerant flow control. The
pressure of the liquid is reduced to the evapor-
ator pressure as the liquid passes through the
refrigerant flow control so that the saturation
temperature of the refrigerant entering the
evaporator will be below the temperature of the
refrigerated space. It will be shown later that a

part of the liquid vaporizes as it passes through
the refrigerant control in order to reduce the
temperature of the liquid to the evaporating
temperature.

In the evaporator, the liquid vaporizes at a
constant pressure and temperature as heat to
supply the latent heat of vaporization passes
from the refrigerated space through the walls of
the evaporator to the vaporizing liquid. By the
action of the compressor, the vapor resulting
from the vaporization is drawn from the
evaporator through the suction line into the
suction inlet of the compressor. The vapor
leaving the evaporator is saturated and its
temperature and pressure are the same as those
of the vaporizing liquid. While flowing through
the suction line from the evaporator to the
compressor, the vapor usually absorbs heat
from the air surrounding the suction line and
becomes superheated. Although the tempera-
ture of the vapor increases somewhat in the
suction line as the result of superheating, the
pressure of the vapor does not change so that

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 81

II
!i

«
£

L-
C V

9- 3
£f

2<5

■i •

82 PRINCIPLES OF REFRIGERATION

Liquid-vapor mixture
30'F-28.46 psig

Liquid-vapor mixture
30T-28.46 psig "

Saturated vapor
30°F-28.46 psig

Superheated vapor
132*^-120.6 psig"

Superheated vapo r t ^ >->

70°F-28.46 psig~Xi£l(£

Subcooled liquid
86°F-120.6 psig

Saturated vapor
102"F-120.6 psig

Liquid-vapor mixture
" 102°F-120.6 psig

Saturated liquid

102°F-120.6 psig

Fig. 6-15. Typical refrigeration system showing the condition of the refrigerant at various points.

the pressure of the vapor entering the compres-
sor is the same as the vaporizing pressure.*

In the compressor, the temperature and
pressure of the vapor are raised by compression
and the high-temperature, high-pressure vapor
is discharged from the compressor into the
hot-gas line. The vapor flows through the
hot-gas line to the condenser where it gives up
heat to the relatively cool air being drawn
across the condenser by the condenser fan.
As the hot vapor gives off heat to the cooler air,
its temperature is reduced to the new saturation
temperature corresponding to its new pressure
and the vapor condenses back into the liquid
state as additional heat is removed. By the
time the refrigerant reaches the bottom of the
condenser, all of the vapor is condensed and the
liquid passes into the receiver tank, ready to be
recirculated.

6-18. The Compression Process. In
modern, high speed compressors, compression
takes place very rapidly and the vapor is in
contact with the compressor cylinder for only a
short time. Because the time of compression
is short and because the mean temperature

* Actually, the pressure of the vapor decreases
slightly between the evaporator and compressor
because of the friction loss in the suction line
resulting from the flow.

differential between the refrigerant vapor and
the cylinder wall is small, the flow of heat
either to or from the refrigerant during com-
pression is usually negligible. Therefore, com-
pression of the vapor in a refrigeration com-
pressor is assumed to occur adiabatically.

Although no heat as such is transferred either
to or from the refrigerant during the compres-
sion, the temperature and enthalpy of the vapor
are increased because of the mechanical work
done on the vapor by the piston. Whenever a
vapor is compressed, unless the vapor is cooled
during the compression, the internal kinetic
energy of the vapor is increased by an amount
equal to the amount of work done on the vapor
(Section 3-12). Therefore, when a vapor is
compressed adiabatically, as in a refrigeration
compressor, wherein no heat is removed from
the vapor during the compression, the tempera-
ture and enthalpy are increased in direct
proportion to the amount of work done during
the compression. The greater the work of
compression, the greater is the increase in
temperature and enthalpy.

The energy equivalent of the work done is
called the heat of compression. The energy to
do the work of compression, which is trans-
ferred to the vapor during the compression
process, is supplied by the compressor driver,

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 83

usually an electric motor. It will be shown
later that the theoretical horsepower required
to drive the compressor can be calculated from
the heat of compression.
6-19. Discharge Temperature. Care should
be taken not to confuse discharge temperature
with condensing temperature. The discharge
temperature is that at which the vapor is dis-
charged from the compressor, whereas the
condensing temperature is that at which the
vapor condenses in the condenser and is the
saturation temperature of the vapor correspond-
ing to the pressure in the condenser. Because
the vapor is usually superheated as it enters the
compressor and because it contains the heat of
compression, the vapor discharged from the
compressor is highly superheated and its
temperature is considerably above the satura-
tion temperature corresponding to its pressure.
The discharge vapor is cooled to the condensing
temperature as it flows through the hot-gas line
and through the upper part of the condenser,
whereupon the further removal of heat from
the vapor causes the vapor to condense at the
saturation temperature corresponding to the
pressure in the condenser.
6-20. Condensing Temperature. To provide
a continuous refrigerating effect the refrigerant
vapor must be condensed in the condenser at
the same rate that the refrigerant liquid is
vaporized in the evaporator. This means that
heat must leave the system at the condenser at
the same rate that heat is taken into the system
in the evaporator and suction line, and in the
compressor as a result of the work of compres-
sion. Obviously, any increase in the rate of
vaporization will increase the required rate of
heat transfer at the condenser.

The rate at which heat will flow through the
walls of the condenser from the refrigerant
vapor to the condensing medium is the function
of three factors: (1) the area of the condensing
surface, (2) the coefficient of conductance of the
condenser walls, and (3) the temperature differ-
ence between the refrigerant vapor and the
condensing medium. For any given condenser,
the area of the condensing surface and the
coefficient of conductance are fixed so that the
rate of heat transfer through the condenser
walls depends only on the temperature differ-
ence between the refrigerant vapor and the
condensing medium.

Since the condensing temperature is always
equal to the temperature of the condensing
medium plus the temperature difference between
the condensing refrigerant and the condensing
medium, it follows that the condensing tempera-
ture varies directly with the temperature of the
condensing medium and with the required rate
of heat transfer at the condenser.
6-21. Condensing Pressure. The condensing
pressure is always the saturation pressure
corresponding to the temperature of the liquid-
vapor mixture in the condenser.

When the compressor is not running, the
temperature of the refrigerant mixture in the
condenser will be the same as that of the
surrounding air, and the corresponding satura-
tion pressure will be relatively low. Conse-
quently, when the compressor is started, the
vapor pumped over into the condenser will not
begin to condense immediately because there is
no temperature differential between the refriger-
ant and the condensing medium, and therefore
no heat transfer between the two. Because of
the throttling action of the refrigerant control,
the condenser may be visualized as a closed
container, and as more and more vapor is
pumped into the condenser without condensing,
the pressure in the condenser increases to a
point where the saturation temperature of
vapor is sufficiently high to permit the required
rate of heat transfer between the refrigerant
and the condensing medium. When the required
rate of heat transfer is reached, the vapor will
condense as fast as it is pumped into the
condenser, whereupon the pressure in the con-
denser will stabilize and remain more or less
constant during the balance of the running cycle.
6-22. Refrigerating Effect. The quantity of
heat that each pound of refrigerant absorbs
from the refrigerated space is known as the
refrigerating effect. For example, when 1 lb of
ice melts it will absorb from the surrounding air
and from adjacent objects an amount of heat
equal to its latent heat of fusion. If the ice melts
at 32° F it will absorb 144 Btu per pound, so that
the refrigerating effect of 1 lb of ice is 144 Btu.

Likewise, when a liquid refrigerant vaporizes
as it flows through the evaporator it will absorb
an amount of heat equal to that required to
vaporize it; thus the refrigerating effect of 1 lb
of liquid refrigerant is potentially equal to its
latent heat of vaporization. If the temperature

84 PRINCIPLES OF REFRIGERATION

of the liquid entering the refrigerant control
from the liquid line is exactly equal to the
vaporizing temperature in the evaporator the
entire pound of liquid will vaporize in the
evaporator and produce useful cooling, in
which case the refrigerating effect per pound
of refrigerant circulated will be equal to the
total latent heat of vaporization. However, in
an actual cycle the temperature of the liquid
entering the refrigerant control is always
considerably higher than the vaporizing tem-
perature in the evaporator, and must first be
reduced to the evaporator temperature before
the liquid can vaporize in the evaporator and
absorb heat from the refrigerated space. For
this reason, only a part of each pound of Hquid
actually vaporizes in the evaporator and
produces useful cooling. Therefore, the refriger-
ating effect per pound of liquid circulated is
always less than the total latent heat of vapori-
zation.

With reference to Fig. 6-15, the pressure of
the vapor condensing in the condenser is 120
psig and the condensing temperature (satur-
ation temperature) of the R-12 vapor corre-
sponding to this pressure is 102° F. Since
condensation occurs at a constant temperature,
the temperature of the liquid resulting from the
condensation is also 102° F. After condensation,
as the liquid flows through the lower part of
the condenser it continues to give up heat to the
cooler condensing medium, so that before the
liquid leaves the condenser its temperature is
usually reduced somewhat below the tempera-
ture at which it condensed. The liquid is then
said to be subcooled. The temperature at
which the liquid leaves the condenser depends
upon the temperature of the condensing
medium and upon how long the hquid remains
in contact with the condensing medium after
condensation.

The liquid may be further subcooled in the
receiver tank and in the liquid line by surrender-
ing heat to the surrounding air. In any case,
because of the heat exchange between the
refrigerant in the liquid line and the surrounding
air, the temperature of the liquid approaching
the refrigerant control is likely to be fairly
close to the temperature of the air surrounding
the liquid line. In Fig. 6-1 5, the liquid ap-
proaches the refrigerant control at a temperature
of 86° F, whereas its pressure is still the same as

the condensing pressure, 120.6 psig. Since the
saturation temperature corresponding to 120.6
psig is 102° F, the 86° F liquid at the refrigerant
control is subcooled 16° F (102 - 86) below its
saturation temperature.

Since the saturation pressure corresponding
to 86° F is 93.2 psig, the R-12 can exist in the
liquid state as long as its pressure is not reduced
below 93.2 psig. However, as the liquid passes
through the refrigerant control its pressure is
reduced from 120.6 psig to 28.46 psig, the
saturation pressure corresponding to the 30°
vaporizing temperature of the refrigerant in the
evaporator. Since the R-12 cannot exist as a
liquid at any temperature above the saturation
temperature of 30° F when its pressure is 28.46
psig, the hquid must surrender enough heat
to cool itself from 86° F to 30° F at the instant
that its pressure is reduced in passing through
the refrigerant control.

From Table 16-3, the enthalpy of hquid at
86° F and at 30° F is 27.73 Btu per pound and
14.76 Btu per pound, respectively, so that each
pound of liquid must surrender 12.97 Btu
(27.73 - 14,76) in order to cool from 86° F to
30° F. Because the hquid expands through the
refrigerant control so rapidly, the liquid is not
in contact with the control for a sufficient length
of time to permit this amount of heat to be
transferred from the refrigerant to the control.
Therefore, a portion of each pound of hquid
vaporizes as the liquid passes through the con-
trol, and the heat to supply the latent heat of
vaporization for the portion that vaporizes is
drawn from the body of the hquid, thereby
reducing the temperature of the refrigerant to the
evaporator temperature. In this instance, enough
of each pound of liquid vaporizes while passing
through the refrigerant control to absorb exactly
the 12.97 Btu of sensible heat that each pound of
hquid must surrender in order to cool from 86° F
to 30° F and the refrigerant is discharged from
the refrigerant control into the evaporator as a
liquid-vapor mixture. Obviously, only the liquid
portion of the liquid-vapor mixture will vaporize
in the evaporator and produce useful cooling.
That portion of each pound of hquid circulated
which vaporizes in the refrigerant control pro-
duces no useful cooling and represents a loss of
refrigerating effect. It follows, then, that the
refrigerating effect per pound of hquid circulated
is equal to the total latent heat of vaporization

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 85

less the amount of heat absorbed by that part of
each pound that vaporizes in the control to
reduce the temperature of the liquid to the
vaporizing temperature.

From Table 16-3, the latent heat of vaporiza-
tion of 'R-12 at 30° F is 66.85 Btu per pound.
Since the loss of refrigerating effect is 12.97 Btu
per pound, the refrigerating effect in this instance
is (66.85 - 12.97) 53.88 Btu per pound.

The percentage of each pound of refrigerant
that vaporizes in the refrigerant control can be
determined by dividing the total latent heat of
vaporization into the heat absorbed by that part
of the pound that vaporizes in the control. In
this instance the percentage of each pound
vaporizing in the control is (12.97/66.85 x 100)
19.4%. Only 80.6% of each pound circulated
actually vaporizes in the evaporator and pro-
duces useful cooling (66.85 x 0.806 = 53.88
Btu/lb).

Even though a portion of each pound cir-
culated vaporizes as it passes through the re-
frigerant control, the enthalpy of the refrigerant
does not change in the control. That is, since
there is no heat transfer between the refrigerant
and the control, the enthalpy of the liquid-vapor
mixture discharged from the control into the
evaporator is exactly the same as the enthalpy of
the liquid approaching the control. Therefore,
the difference between the enthalpy of the re-
frigerant vapor leaving the evaporator and the
enthalpy of the liquid approaching the control is
only the amount of heat absorbed by the re-
frigerant in the evaporator, which is, of course,
the refrigerating effect. Hence, for any given
conditions the refrigerating effect per pound can
be easily determined by subtracting the enthalpy
of the liquid refrigerant entering the control
from the enthalpy of the saturated vapor leaving
the evaporator.

Example 6-1. Determine the refrigerating
effect per pound if the temperature of the liquid
R-12 approaching the refrigerant control is
86° F and the temperature of the saturated
vapor leaving the evaporator is 30° F.

Solution. From Table
16-3, enthalpy of R-12 satur-
ated vapor at 30° F = 81.61 Btu/lb

Enthalpy of R-12 liquid
at 86° F = 27.73 Btu/lb

Refrigerating effect per
pound = 53.88 Btu/lb

Example 6-2. If, in Example 6-1, the tem-
perature of the liquid entering the refrigerant
control is 60° F rather than 86° F, determine the
refrigerating effect.

Solution. From Table
16-3, enthalpy of R-12
saturated vapor at 30° F =81.61 Btu/lb

Enthalpy of R-12 liquid
at 60° F = 21.57 Btu/lb

Refrigerating effect = 60.04 Btu/lb

Example 6-3. If, in Example 6-1, the
pressure in the evaporator is 21.05 psig, and the
liquid reaching the refrigerant control is 86°" F,
what is the refrigerating effect?

Solution. From Table
16-3, the saturation tem-
perature of R-12 corre-
sponding to 21.05 psig is
20° F and the enthalpy of
R-12 saturated vapor at that
temperature = 80.49 Btu/lb

Enthalpy of R-12 liquid
at 86° F = 27.73 Btu/lb

Refrigerating effect = 52.77 Btu/lb

A comparison of Examples 6-1 and 6-2 in-
dicates that the refrigerating effect increases as
the temperature of the liquid approaching the
refrigerant control decreases, whereas a com-
parison of Example 6-1 and 6-3 shows that the
refrigerating effect decreases as the vaporizing
temperature decreases. Therefore, it is evident
that the refrigerating effect per pound of liquid
circulated depends upon two factors: (1) the
evaporating temperature and (2) the temperature
at which the liquid refrigerant enters the re-
frigerant control. The higher the evaporating
temperature and the lower the temperature of
the liquid entering the refrigerant control,
the greater will be the refrigerating effect.
6-23. System Capacity. The capacity of any
refrigerating system is the rate at which it will
remove heat from the refrigerated space and is
usually stated in Btu per hour or in terms of its
ice-melting equivalent.

Before the era of mechanical refrigeration, ice
was widely used as a cooling medium. With the
development of mechanical refrigeration, it was
only natural that the cooling capacity of
mechanical refrigerators should be compared
with an ice-melting equivalent.

When one ton of ice melts.it will absorb
288,000 Btu (2000 lb x 144 Btu/lb). If one ton

86 PRINCIPLES OF REFRIGERATION

of ice melts in one day (24 hr), it will absorb heat
at the rate of 12,000 Btu/hr (288,000 Btu/24 hr)
or 200 Btu/min (12,000 Btu/hr/60). Therefore,
a mechanical refrigerating system having the
capacity of absorbing heat from the refrigerated
space at the rate of 200 Btu/min (12,000 Btu/hr)
is cooling at a rate equivalent to the melting of
one ton of ice in 24 hr and is said to have a
capacity of one ton.

The capacity of a mechanical refrigeration
system, that is, the rate at which the system will
remove heat from the refrigerated space, depends
upon two factors: (1) the weight of refrigerant
circulated per unit of time and (2) the refrigerat-
ing effect of each pound circulated.

Example 6-4. A mechanical refrigerating
system is operating under conditions such that
the vaporizing temperature is 30° F while the
temperature of the liquid approaching the re-
frigerant control is 86° F. If R-12 is circulated
through the system at the rate of 5 lb/min,
determine:

(«) the refrigerating capacity of the system in
Btu per hour.

(b) the refrigerating capacity of the system in
tons.

Solution

(a) From Example 6-1,

refrigerating effect
Weight of refrigerant

circulated per minute
Refrigerating capacity

in Btu per minute

Refrigerating capacity
in Btu per hour

(b) Refrigerating capacity

in tons

= 53.88 Btu/lb

= 5 lb

= 5 x 53.88

= 269.40 Btu/min

= 269.40 x 60
= 16,164 Btu/hr

_ 269.40

200
= 1.347 tons

6-24. Weight of Refrigerant Circulated per
Minute per Ton. The weight of refrigerant
which must be circulated per minute per ton of
refrigerating capacity for any given operating
conditions is found by dividing the refrigerating
effect per pound at the given conditions into 200.

Example 6-5. An R-12 system is operating
at conditions such that the vaporizing tempera-
ture is 20° F and the condensing temperature is
100° F. If it is assumed that no subcooling of

the liquid occurs so that the temperature of the
liquid at the refrigerant control is also 100° F,
find:

(a) The refrigerating effect per pound

(b) The weight of refrigerant circulated per
minute per ton

Solution

(a) From Table 16-3, en-

thalpy of R-12 satur-
ated vapor at 20° F = 80.49 Btu/lb
Enthalpy of R-12

liquid at 100° F = 31.16 Btu/lb

Refrigerating effect 49.33 Btu/lb

(b) Weight of refrigerant cir-

culated per minute per

ton 200

49.33
4.05 lb

10 x 4.05
40.5 lb

Example 6-6. If, in Example 6-5, the liquid
is subcooled from 100° F to 80° F before it
reaches the refrigerant control, calculate:

(a) the refrigerating effect

(b) the weight of refrigerant circulated per
minute per ton

Solution

(a) From Table 16-3,

enthalpy of R-12

saturated vapor at

20° F
Enthalpy of R-12 liquid

at 80° F
Refrigerating effect

(b) Weight of refrigerant cir-

culated per minute per
ton

= 80.49 Btu/lb

= 26J8 Btu/lb
= 54.21 Btu/lb
200

54.21
= 3.69 lb

In comparing Examples 6-5 and 6-6, it is
apparent that the weight of refrigerant which
must be circulated per minute per ton of re-
frigerating capacity varies with the refrigerating
effect and depends upon the operating conditions
of the system. As the refrigerating effect per
pound increases, the weight of refrigerant cir-
culated per minute per ton decreases.
6-25. Volume of Vapor Displaced per Min-
ute per Ton. When 1 lb of liquid refrigerant
vaporizes, the volume of vapor which results

REFRIGERATION AND THE VAPOR COMPRESSION SYSTEM 87

depends upon the vaporizing temperature. The
lower the vaporizing temperature and pressure,
the greater is the volume of the vapor which is
produced. When the vaporizing temperature is
known, the specific volume of the saturated
vapor which results from the vaporization can
be found in the saturated vapor tables. For
instance, from Table 16-3, the specific volume of
R-12 saturated vapor at 10° F is 1.351 cu ft per
pound. This means that each pound of R-12
that vaporizes at 10° F produces 1.351 cuft of
vapor. Therefore, if 10 lb of R-12 are vaporized
at 10° F in an evaporator each minute, saturated
vapor will be produced at the rate of 13.51 cu ft
per minute (10 x 1.351).

In order to produce one ton of refrigerating
capacity, a definite weight of refrigerant must be
vaporized each minute. The volume of vapor
which must be removed from the evaporator
each minute can be calculated by multiplying
the weight of refrigerant circulated per minute
by the specific volume of the saturated vapor at
the vaporizing temperature.

Example 6-7. Determine the volume of
vapor to be removed from the evaporator per
minute per ton of refrigerating capacity for the
system described in Example 6-5.

Solution. From
Table 16-3, specific
volume of R-12 satur-
ated vapor at 20° F

From Example 6-5,
weight of refrigerant
circulated per minute
per ton

Volume of vapor
displaced per minute
per ton

= 1.121 cu ft/lb

= 4.05 lb/min/ton

= 4.05 x 1.121

= 4.55 cu ft/min/ton

6-26. Compressor Capacity. In any mecha-
nical refrigerating system the capacity of the
compressor must be such that vapor is drawn
from the evaporator at the same rate that vapor
is produced by the boiling action of the liquid
refrigerant. If the refrigerant vaporizes faster
than the compressor is able to remove the vapor,
the excess vapor will accumulate in the evapora-
tor and cause the pressure in the evaporator to
increase, which in turn will result in raising the
boiling temperature of the liquid. On the other
hand, if the capacity of the compressor is such
that the compressor removes the vapor from the

evaporator too rapidly, the pressure in the
evaporator will decrease and result in a decrease
in the boiling temperature of the liquid. In either
case, design conditions will not be maintained
and the refrigerating system will be unsatis-
factory.

The maintenance of design conditions and
therefore good refrigeration depends upon the
selection of a compressor whose capacity is such
that the compressor will displace in any given in-
terval of time a volume of vapor that is equal to
the volume occupied by the weight of refrig-
erant which must be vaporized during the same
time interval in order to produce the required
refrigerating capacity at the design conditions.

For instance, in Example 6-7, 4.05 lb of R-12
must be vaporized each minute at 20° F for each
one ton of refrigerating capacity desired. In
vaporizing, the 4.05 lb of R-12 produce 4.55
cuft of vapor (4.05 x 1.121). If the evaporator
pressure and the boiling temperature of the
liquid in the evaporator are to remain constant,
this volume of vapor must be removed from the
evaporator each minute for each one ton of
refrigerating capacity. Hence, the compressor
selected for a system operating at the conditions
of Example 6-7 should have a capacity such that
it will remove vapor from the evaporator at the
rate of 4.55 cu ft per minute for each ton of
refrigerating capacity required. For a 10 ton
system, the compressor would have to remove
vapor from the evaporator at the rate of 45.50
cu ft per minute (10 x 4.55).

PROBLEMS

1. The temperature of liquid R-12 entering the
refrigerant control is 86° F and the vaporizing
temperature 30° F. Determine:

(a) The refrigerating effect per pound of
refrigerant circulated. Ans. 53.89 Btu/lb

(6) The loss of refrigerating effect per pound.

Ans. 12.96 Btu/lb

(d)The volume of vapor displaced per
minute per ton. Ans. 3.48 cu ft/min/ton

2. If saturated R-12 liquid reaches the refriger-
ant control at a pressure of 136 psig and the
vaporizing pressure in the evaporator is 30.07
psig, determine:

(a) The refrigerating effect per pound.

Ans. 48.18 Btu/lb

88 PRINCIPLES OF REFRIGERATION

(*) The weight of refrigerant circulated per (c) The volume of vapor to be displaced per

minute per ton. Arts. 4.15 lb/min/ton minute per ton. Arts. 3.13 cuft/min/ton

per ton. Arts. 3.77 cu ft/min/ton 70° F and the evaporating pressure is lowered to

3. If the liquid approaching the refrigerant 16.35 psig, determine

control in Problem 2 is subcooled to 70° F, (a) The refrigerating effect. Arts. 45.25 Btu/lb

determine: (b) The weight of refrigerant circulated per

(a) The refrigerating effect. Arts. 57.93 Btu/lb minute per ton. Arts. 4.42 lb/min/ton

(b) The weight of refrigerant circulated per (c) The volume of vapor displaced per
minute per ton. Arts. 3.45 cu ft/min/ton minute per ton. Arts. 6.44 cu ft/min/ton

Cycle Diagrams
and the Simple
Saturated Cycle

7-1. Cycle Diagrams. A good knowledge of
the vapor-compression cycle requires an inten-
sive study not only of the individual processes
that make up the cycle but also of the relation-
ships that exist between the several processes and
of the effects that changes in any one process in
the cycle have on all the other processes in the
cycle. This is greatly simplified by the use of
charts and diagrams upon which the complete
cycle may be shown graphically. Graphical
representation of the refrigeration cycle permits
the desired simultaneous consideration of all the
various changes in the condition of the re-
frigerant which occur during the cycle and the
effect that these changes have on the cycle with-
out the necessity of holding in mind all the
different numerical values involved in cyclic
problems.

The diagrams frequently used in the analysis of
the refrigeration cycle are the pressure-enthalpy
(Ph) diagram and the temperature-entropy (Ts)
diagram. Of the two, the pressure-enthalpy
diagram seems to be the most useful and is the
one which is emphasized in the following sec-
tions. The temperature-entropy diagram has
already been introduced (Section 4-19) and its
application to the refrigeration cycle will be
discussed to some extent in this chapter.
7-2. The Pressure-Enthalpy Diagram. A
pressure-enthalpy chart for R-12 is shown in

Fig. 7-1.* The condition of the refrigerant in
any thermodynamic state can be represented as
a point on the Ph chart. The point on the Ph
chart which represents the condition of the
refrigerant in any one particular thermodynamic
state may be located if any two properties of
the refrigerant at that state are known. Once the
state point has been located on the chart, all the
other properties of the refrigerant for that state
can be determined directly from the chart.

As shown by the skeleton Ph chart in Fig.
7-2, the chart is divided into three areas which
are separated from each other by the saturated
liquid and saturated vapor curves. The area
on the chart to the left of the saturated liquid
curve is called the subcooled region. At any
point in the subcooled region the refrigerant is
in the liquid state and its temperature is below
the saturation temperature corresponding to
its pressure. The area to the right of the satu-
rated vapor curve is the superheated region and
the refrigerant is in the form of a superheated
vapor. The center section of the chart, between
the saturated liquid and saturated vapor curves,
represents the change in phase of the refrigerant
between the liquid and vapor states. At any
point between the two curves the refrigerant is
in the form of a liquid-vapor mixture. The
distance between the two curves along any
constant pressure line, as read on the enthalpy
scale at the bottom of the chart, is the latent heat
of vaporization of the refrigerant at that
pressure. The saturated liquid and saturated
vapor curves are not exactly parallel to each
other because the latent heat of vaporization
of the refrigerant varies with the pressure at
which the change in phase occurs.

On the chart, the change in phase from the
liquid to the vapor phase takes place progres-
sively from left to right, whereas the change in
phase from the vapor to the liquid phase occurs
from right to left. Close to the saturated liquid
curve the liquid-vapor mixture is nearly all
liquid, whereas close to the saturated vapor
curve the liquid-vapor mixture is almost all
vapor. The lines of constant quality (Fig. 7-3),
extending from top to bottom through the
center section of the chart and approximately
parallel to the saturated liquid and vapor

* The pressure-enthalpy chart for each refrigerant
is different, depending upon the properties of the
particular refrigerant.

90 PRINCIPLES OF REFRIGERATION

(eisd) ajnsssjd

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 91

Subcooled region
(Refrigerant is in
the form of a
subcooled liquid)

Region of phase change
(Refrigerant is a liquid-
vapor mixture)

>— Liquid to vapor > >

Vapor to liquid — < «—

Superheated region
(Refrigerant is in
the form of a
superheated vapor)

Saturated liquid curve

Saturated vapor curve-

Specific enthalpy (Btu per lb)
Fig. 7-2. Skeleton Mi chart Illustrating the three regions of the chart and the direction of phase changing.

curves, indicate the percentage of vapor in the
mixture in increments of 10%. For example, at
any point on the constant quality line closest to
the saturated liquid curve the quality of the
liquid-vapor mixture is 10%, which means that
10% (by weight) of the mixture is vapor.
Similarly, the indicated quality of the mixture
at any point along the constant quality line
closest to the saturated vapor curve is 90% and
the amount of vapor in the liquid-vapor mixture
is 90%. At any point on the saturated liquid
curve the refrigerant is a saturated liquid and at
any point along the saturated vapor curve the
refrigerant is a saturated vapor.

The pressure is plotted along the vertical
axis, and the enthalpy is plotted along the
horizontal axis. Hence, the horizontal lines
extending across the chart are lines of constant
pressure and the vertical lines are lines of
constant enthalpy.

The lines of constant temperature in the
subcooled region are almost vertical on the
chart and paralled to the lines of constant
enthalpy. In the center section, since the
refrigerant changes state at a constant tempera-
ture and pressure, the lines of constant tempera-
ture run horizontally across the chart and
parallel to the lines of constant pressure. At the
saturated vapor curve the lines of constant
temperature change direction again and, in the

superheated vapor region, fall off sharply
toward the bottom of the chart.

The straight lines which extend diagonally
and almost vertically across the superheated
vapor region are lines of constant entropy. The
curved, nearly horizontal lines crossing the
superheated vapor region are lines of constant
volume.

The values of any of the various properties
of the refrigerant which are of importance in the
refrigerating cycle may be read directly from
the Ph chart at any point where the value of that
particular property is significant to the process
occurring at that point. To simplify the chart,
the number of lines on the chart is kept to a
minimum. For this reason, the value of those
properties of the refrigerant which have no real
significance at some points in the cycle are
omitted from the chart at these points. For
example, in the liquid region and in the region
of phase change (center section) the values of
entropy and volume are of no particular
interest and are therefore omitted from the
chart in these sections.

Since the Ph chart is based on a 1 lb mass of
the refrigerant, the volume given is the specific
volume, the enthalpy is in Btu per pound, and
the entropy is in Btu per pound per degree of
absolute temperature. Enthalpy values are
found on the horizontal scale at the bottom of

92 PRINCIPLES OF REFRIGERATION

the chart and the values of entropy and volume
are given adjacent to the entropy and volume
lines, respectively. The values of both enthalpy
and entropy are based on the arbitrarily selected
zero point of —40° F.

The magnitude of the pressure in psia is
read on the vertical scale at the left side of the
chart. Temperature values in degrees Fahren-
heit are found adjacent to the constant tempera-
ture lines in the subcooled and superheated
regions of the chart and on both the saturated
liquid and saturated vapor curves.
7-3. The Simple Saturated Refrigerating
Cycle. A simple saturated refrigerating cycle is
a theoretical cycle wherein it is assumed that the
refrigerant vapor leaves the evaporator and
enters the compressor as a saturated vapor (at
the vaporizing temperature and pressure) and
the liquid leaves the condenser and enters the
refrigerant control as a saturated liquid (at
the condensing temperature and pressure).
Although the refrigerating cycle of an actual
refrigerating machine will usually deviate
somewhat from the simple saturated cycle, the
analysis of a simple saturated cycle is nonethe-
less worthwhile. In such a cycle, the funda-
mental processes which are the basis of every
actual vapor compression refrigerating cycle are
easily identified and understood. Furthermore,

by using the simple saturated cycle as a standard
against which actual cycles may be compared,
the relative efficiency of actual refrigerating
cycles at various operating conditions can be
readily determined.

A simple saturated cycle for a R-12 system is
plotted on a Ph chart in Fig. 7-4. The system is
assumed -to be operating under such conditions
that the vaporizing pressure in the evaporator is
35.75 psia and the condensing pressure in the
condenser is 131.6 psia. The points A, B, C, D,
and E on the Ph diagram correspond to points
in the refrigerating system as shown on the flow
diagram in Fig. 7-5.

At point A, the refrigerant is a saturated
liquid in the condenser at the condensing
pressure and temperature, and its properties, as
given in Table 16-3, are:

p = 131.6 psia t = 100° F

h = 31.16 Btu/lb s = 0.06316 Btu/lb/° F
v = 0.0127 cu ft/lb

At point A, the values of p, t, and h may be
read directly from the Ph chart. Since the
refrigerant is always a saturated liquid at point
A, point A will always fall somewhere along
the saturated liquid curve and can be located on
the Ph chart if either/?, t, or h is known. Usually

Specific enthalpy (Btu per lb)

Fig. 7-3. Skeleton Ph chart showing paths of constant pressure, constant temperature, constant volume,
constant quality, constant enthalpy, and constant entropy. (Refrigerant- 1 2.)

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 93

h x h a he h e hd

Specific enthalpy (Btu per lb)
Fig. 7-4. Pressure-enthalpy diagram of a simple saturated cycle operating at a vaporizing temperature of 20° F
and a condensing temperature of 100° F. (Refrigerant- 1 2.)

in actual practice, either p, t, or both will be
measurable.

7-4. The Expansion Process. In the simple
saturated cycle there is assumed to be no change
in the properties (condition) of the refrigerant
liquid as it flows through the liquid line from
the condenser to the refrigerant control and
the condition of the liquid approaching the re-
frigerant control is the same as its condition
at point A. The process described by the initial
and final state points A-B occurs in the refriger-
ant control when the pressure of the liquid
is reduced from the condensing pressure to the
evaporating pressure as the liquidpassesthrough
the control.* When the liquid is expanded into
the evaporator through the orifice of the control,
the temperature of the liquid is reduced from the
condensing temperature to the evaporating tem-
perature by the flashing into vapor of a small
portion of the liquid.
Process A-B is a throttling type of adiabatic

* Process A-B is an irreversible adiabatic ex-
pansion during which the refrigerant passes through
a series of state points in such a way that there is no
uniform distribution of any of the properties.
Hence, no true path can be drawn for the process and
line A-B merely represents a process which begins
at state point A and terminates at state point B.

expansion, frequently called "wire-drawing,"
in which the enthalpy of the working fluid does
not change during the process. This type of
expansion occurs whenever a fluid is expanded
through an orifice from a high pressure to a
lower pressure. It is assumed to take place
without the gain or loss of heat through the
piping or valves and without the performance of

work.f

Since the enthalpy of the refrigerant does not
change during process A-B, point B is located
on the Ph chart by following the line of constant
enthalpy from point A to the point where the
constant enthalpy line intersects the line of
constant pressure corresponding to the evapor-
ating pressure. To locate point B on the Ph
chart, the evaporating pressure or temperature
must be known.

As a result of the partial vaporization of
the liquid refrigerant during process A-B, the

t Actually, a certain amount of work is done by
the fluid in projecting itself through the orifice of
the control. However, since the heat equivalent
of the work done in overcoming the friction of the
orifice merely heats the orifice and is subsequently
reabsorbed by the fluid, the assumption that the
enthalpy of the fluid does not change during the
process is not in error.

94 PRINCIPLES OF REFRIGERATION

refrigerant at point B is a liquid-vapor mixture
whose properties are :
p = 35.75 psia
t = 20° F

h =31.16 Btu/lb (same as at point A)
v = 0.1520 cu ft/lb
s - 0.06316 Btu/lb/°F

Note. The change in entropy during the
process A-B results from a transfer of heat
energy which takes place within the refrigerant
itself because of internal friction. A transfer of
energy which occurs entirely within the working
fluid does not affect the enthalpy of the fluid,
only the entropy changes.

At point B, in addition to the values of p, t,
and h, the approximate quality of the vapor can
be determined from the Ph chart by interpolat-
ing between the lines of constant quality. In
this instance, the quality of the vapor as deter-
mined from the Ph chart is approximately
27%.

Since the refrigerant at point B is a liquid-
vapor mixture, only the values of p and t can be
read directly from Table 16-3. However,
because the enthalpy of the refrigerant at points
A and B is the same, the enthalpy at point B
may be read from Table 16-3 as the enthalpy
at the conditions of point A. The quality of the
vapor at point B can be determined as in Section
6-22, using enthalpy values taken either from
Table 16-3 or from the Ph chart directly.

The values of s and v at point B are usually of
no interest and are not given either on the Ph
chart or in the vapor tables. If the values of s
and v are desired, they must be calculated.
7-5. The Vaporizing Process. The process
B-C is the vaporization of the refrigerant in the
evaporator. Since vaporization takes place at a
constant temperature and pressure, B-C is both
isothermal and isobaric. Therefore, point C
is located on the Ph chart by following the lines
of constant pressure and constant temperature
from point B to the point where they intersect
the saturated vapor curve. At point C the
refrigerant is completely vaporized and is a
saturated vapor at the vaporizing temperature
and pressure. The properties of the refrigerant
at point C, as given in Table 16-3 or as read
from the Ph chart, are:

p •* 35.75 psia (same as at point B)

t = 20° F (same as at point B)

h = 80.49 Btu/lb

v =• 1.121 cu ft/lb

s = 0.16949 Btu/lb/°F
The enthalpy of the refrigerant increases
during process B-C as the refrigerant flows
through the evaporator and absorbs heat from
the refrigerated space. The quantity of heat
absorbed by the refrigerant in the evaporator
(refrigerating effect) is the difference between
the enthalpy of the refrigerant at points B and
C. Thus, if h a , h b , h , h d , h e , and h x represent the

Point at which
vaporization isv£~~
complete ^>

Suction vapor flows
from the evaporator
to the compressor
through the suction
line without a
change in condition

Discharge vapor
'from compressor

Refrigerant after
passing through
refrigerant control

Point at which
condensation-
begins

In the simple saturated
cycle, the refrigerant
flows through the liquid
line from the condenser
to the refrigerant
control without a
change in condition

/-S

£l

Point at which r
condensation is—'
complete

Fig. 7-5. Flow diagram of a
simple saturated cycle.

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 95

enthalpies of the refrigerant at points A,B,C,D,
E, and X, respectively, then

qi=h c - K (7-1)

where q x = the refrigerating effect in Btu/lb.
But since h t is equal to h a , then

^ = A. - K (7-2)

When we substitute the appropriate values in
Equation 7-2 for the example in question,

ft = 80.49 - 31.16

= 49.33 Btu/lb

On the Ph diagram, the distance between
point X and point C represents the total latent
heat of vaporization of 1 lb of R-12 at the
vaporizing pressure of 35.75 psia (h fg in Table
16-3). Therefore, since the distance B-C is the
useful refrigerating effect, the difference between
X-C and B-C, which is the distance X-B, is
the loss of refrigerating effect.
7-6. The Compression Process. In the simple
saturated cycle, the refrigerant undergoes no
change in condition while flowing through the
suction line from the evaporator to the com-
pressor. Process C-D takes place in the
compressor as the pressure of the vapor is
increased by compression from the vaporizing
pressure to the condensing pressure. For the
simple saturated cycle, the compression process,
C-D, is assumed to be isentropic* An isen-
tropic compression is a special type of adiabatic
process which takes place without friction, f
It is sometimes described as a "frictionless-
adiabatic" or "constant-entropy" compression.

According to Equation 4-3, Section 4-19, the
change in entropy (As) during any process is
equal to the transferred heat (Ag) divided by the
average absolute temperature ( C R). In any
frictionless-adiabatic process, such as the com-

* It will be shown later that compression of the
vapor in an actual refrigerating compressor usually
deviates somewhat from true isentropic compression.
As a general rule, compression is polytropic.

t The term, adiabatic, is used to describe any
number of processes which take place without the
transfer of energy as heat to or from the working
substance during the process. Thus, an isentropic
process is only one of a number of different processes
which may be termed adiabatic. For example,
compare process C-D with process A-B. Both are
adiabatic, but C-D is frictionless, whereas A-B is
a throttling type of process which involves friction.

pression process C-D, wherein no heat, as such,
is transferred either internally (within the vapor
itself) or externally (to or from an external
source) Ag will always be equal to zero. If Ag
is equal to zero, then As must also be equal to
zero. Hence, there is no change in the entropy
of the vapor during a frictionless-adiabatic
(isentropic) compression.

Since there is no change in the entropy of the
vapor during process C-D, the entropy of the
refrigerant at point D is the same as at point &
Therefore, point D can be located on the Ph
chart by following the line of constant entropy
from point C to the point where the constant
entropy line intersects the hne of constant pres-
sure corresponding to the condensing pressure.

At point D, the refrigerant is a superheated
vapor whose properties are :

p = 131.6 psia

t = 112° F (approximate)

h = 90.6 Btu/lb (approximate)

v = 0.330 cu ft/lb (approximate)

s = 0.16949 Btu/lb/° F (same as at point C)

All of the properties of the refrigerant at the
condition of point D are taken from the Ph
chart. Since the values of t, h, and v require
interpolation, they are only approximations.
The properties of the superheated refrigerant
vapor cannot usually be read accurately from
the vapor table unless the pressure of the vapor
in question corresponds exactly to one of the
pressures listed in the table, This is seldom the
case, particularly at the higher pressures where
the pressure listings in the table are in 10 lb
increments.

Work is done on the vapor during the com-
pression process, C-D, and the enthalpy of the
refrigerant is increased by an amount equal to
the heat energy equivalent of the mechanical
work done on the vapor. The heat energy
equivalent of the work done during the com-
pression is often referred to as the heat of
compression and is equal to the difference in the
enthalpy of the refrigerant at points D and C.
Thus, where q t is the heat of compression per
pound of refrigerant circulated,

q t =h a - h e (7-3)

For the example in question,

q 2 = 90.60 - 80.49
= 10.11 Btu/lb

PRINCIPLES OF REFRIGERATION

The mechanical work done on the vapor
by the piston during the compression may be
calculated from the heat of compression. If
h> is the work done in foot-pounds per pound of
refrigerant circulated and / is the mechanical
energy equivalent of heat, then

w = ft x /

(7-4)

or w = J(h a - hj

(7-5)

when we substitute in Equation 7-4,

w = 10.11 x 778

- 7865.58 ft-lb

As a result of absorbing the heat of com-
pression, the hot vapor discharged from the
compressor is in a superheated condition, that
is, its temperature is greater than the saturation
temperature corresponding to its pressure.
In this instance, the vapor leaves the compressor
at a temperature of 1 12° F, whereas the satura-
tion temperature corresponding to its pressure
of 131.6 psia is 100° F. Thus, before the vapor
can be condensed, the superheat must be
removed and the temperature of the vapor
lowered from the discharge temperature to the
saturation temperature corresponding to its
pressure. *

7-7. The Condensing Process. % Usually, both
processes D-E and E-A take place in the
condenser as the hot gas discharged from the
compressor is cooled to the condensing tempera-
ture and condensed. Process D-E occurs in the
upper part of the condenser and to some extent
in the hot gas line. It represents the cooling of
the vapor from the discharge temperature to the
condensing temperature as the vapor rejects
heat to the condensing medium. During
process D-E, the pressure of the vapor remains
constant and point E is located on the Ph chart
by following a line of constant pressure from
point D to the point where the constant pressure
line intersects die saturated vapor curve.

At point E, the refrigerant is a saturated vapor
at the condensing temperature and pressure.
Its properties, as read from either the Ph chart
or from Table 16-3, are:

p = 131.6 psia (same as at point D)
t = 100° F
h = 88.62 Btu/lb
s - 0.16584 Btu/lb/°F
»» 0.319 cu ft/lb

The quantity of sensible heat (superheat)
removed from 1 lb of vapor in the condenser in
cooling the vapor from the discharge tempera-
ture to the condensing temperature is the differ-
ence between the enthalpy of the refrigerant at
point D and the enthalpy at point E (h a — h t ).

Process E-A is the condensation of the vapor
in the condenser. Since condensation takes
place at a constant temperature and pressure,
process E-A follows along lines of constant
pressure and temperature from point E to
point A. The heat rejected to the condensing
medium during process E-A is the difference
between the enthalpy of the refrigerant at
points E and A (h e — h a ).

On returning to point A, the refrigerant has
completed one cycle and its properties are the
same as those previously described for point A.

Since both processes D-E and E-A occur in
the condenser, the total amount of heat rejected
by the refrigerant to the condensing medium in
the condenser is the sum of the heat quantities
rejected during processes D-E and D-A. The
total heat given up by the refrigerant at the
condenser is the difference between the enthalpy
of the superheated vapor at point D and the
saturated liquid at point A. Hence,

?s = h ~ K (7-6)

where ft = the heat rejected at the condenser
per pound of refrigerant circulated.
In this instance,

ft = 90.60 - 31.16
- 59.44 Btu/lb

If the refrigerant is to reach point A at the
end of the cycle in the same condition as it left
point A at the beginning of the cycle, the total
heat rejected by the refrigerant to the condens-
ing medium in the condenser must be exactly
equal to the heat absorbed by the refrigerant at
all other points in the cycle. In a simple
saturated cycle, the refrigerant is heated at only
two points in the cycle: (1) in the evaporator
by absorbing heat from the refrigerated space
(ft) and in the compressor by the heat of com-
pression (ft).

Therefore,

ft - ft + ft (7-7)

In this instance,

ft =49.33 +10.11

= 59.44 Btu/lb

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 97

Where m is the weight of refrigerant to be
circulated per minute per ton,

200 Btu/min

(7-8)

For the cycle in question,
200

= 4.05 lb/min/ton

Then, where Q a is the total quantity of heat
rejected at the condenser per minute per ton,

Qa = »<fc) (7-9)

or ft = mUhi ~ A„) (7-10)

For the cycle in question,

ft = 4.05 x 59.44

= 240.93 Btu/min/ton

Where Q % is the heat of compression per
minute per ton of refrigerating capacity,

e a =/w(? 8 ) (7-11)

or ft = m(fi d - A,) (7-12)

Substituting,

ft =4.05 x 10.11
= 40.95 Btu/min/ton

Where W is the work of compression done on
the vapor per minute per ton of refrigerating
capacity,

W = rriiw) (7-13)

or, since w equals J(qJ or /(A„ - A„),

W = JmUqd (7-14)

or W = Jm(h d - h£ (7-15)

or W = /(ft) (7-16)

When we substitute in Equation 7-15,

W = 778 x 4.05 x (90.60 - 80.49)

= 31,856 ft-lb/min/ton

7-8. Theoretical Horsepower. The theoreti-
cal horsepower required to drive the compressor
per ton of refrigerating capacity may be found
by applying Equation 1-5 (Section 1-11):

31,856
P ~ 33,000 x 1
= 0.965 hp/ton

A more convenient method of determining
the theoretical horsepower per ton is produced
by combining Equations 1-5 and 7-15:

hp =

m(h d — Aj)
42.42

(7-17)

The compressor horsepower as calculated
above represents only the horsepower required
to compress the vapor. That is, it is the
theoretical power which would be required per
ton of refrigerating capacity by a 100% efficient
system. It does not take into account the power
required to overcome friction in the compressor
and other power losses. The actual (brake)
horsepower required per ton of refrigeration
will usually be from 30% to 50% more than
the theoretical horsepower calculated, depend-
ing upon the efficiency of the compressor.
The factors governing compressor efficiency are
discussed later.

7-9. Coefficient of Performance. The coeffi-
cient of performance of a refrigerating cycle is
an expression of the cycle efficiency and is
stated as the ratio of the heat absorbed in the
refrigerated space to the heat energy equivalent
of the energy supplied to the compressor, that is,

Heat absorbed from
Coefficient of _ the refrigerated space
performance Heat energy equivalent
of the energy supplied
to the compressor

For the theoretical simple saturated cycle, this
may be written as

Refrigerating effect

c.o.p. =

r Heat of compression

(A, - kg)

= (A d - A„)

(ft)
Hence, for the cycle in question,
49.33

(7-18)

c.o.p.

10.11
4.88

7-10. Effect of Suction Temperature on
Cycle Efficiency. The efficiency of the vapor-
compression refrigerating cycle varies consider-
ably with both the vaporizing and condensing

PRINCIPLES OF REFRIGERATION

h a h c h c ' h e hd hd

Specific enthalpy (Btu/lb above - 40° F)
Fig. 7-4. Comparison of two simple saturated cycles operating at different vaporizing temperatures (figure
distorted). (Refrigerant- 12.)

temperatures. Of the two, the vaporizing
temperature has by far the greater effect.

To illustrate the effect that varying the suction
temperature has on cycle efficiency, cycle
diagrams of two simple saturated cycles operat-
ing at different suction temperatures are drawn
on the Ph chart in Fig. 7-6. One cycle, identified
by the points A, B, C, D, and E, is operating at a
vaporizing temperature of 10° F and a condens-
ing temperature of 100° F. A similar cycle
having the same condensing temperature but
operating at a vaporizing temperature of 40° F
is set off by the points A, W, C", £>', and E.

To facilitate a comparison of the two cycles,
the following values have been determined from
the PA chart:

(a) For the 10° F cycle,
qi=h c -h a = 79.36 - 31.16 = 48.20 Btu/lb
q i =h d -h e = 90.90 - 79.36 = 11.54 Btu/lb
q a "h d -h a = 90.90 - 31.16 = 59.74 Btu/lb

(6) For the 40° F cycle,
qi = h c . -h a = 82.71 - 31.16 = 51.55 Btu/lb
q a = h* - A < = 90.20 - 82.71 = 7.49 Btu/lb
fl 8 = h, -h a = 90.20 - 31.16 = 59.04 Btu/lb

In comparing the two cycles, note that the
refrigerating effect per pound of refrigerant
circulated is greater for the cycle having the
higher vaporizing temperature. The refrigerat-

ing effect for the cycle having the 10° F vaporiz-
ing temperature is 48.20 Btu/lb. When the
vaporizing temperature of the cycle is raised to
40° F, the refrigerating effect increases to 51.55
Btu/lb. This represents an increase in the
refrigerating effect per pound of

(he - K) - {h c - h a )
h c — h a

51.55 - 48.20

x 100

48.20
= 6.95%

The greater refrigerating effect per pound of
refrigerant circulated obtained at the higher
vaporizing temperature is accounted for by the
fact that there is a smaller temperature differen-
tial between the vaporizing temperature and the
temperature of the liquid approaching the
refrigerant control. Hence, at the higher
suction temperature, a smaller fraction of the
refrigerant vaporizes in the control and a greater
portion vaporizes in the evaporator and pro-
duces useful cooling.

Since the refrigerating effect per pound is
greater, the weight of refrigerant which must be
circulated per minute per ton of refrigerating
capacity is less at the higher suction temperature
than at the lower suction temperature. Whereas

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 99

the weight of refrigerant circulated per minute
per ton for the 10° F cycle is

200

Ac — f>a

200
= 48.20
= 4.151b/min

The weight of refrigerant circulated per minute
per ton for the 40° F cycle is only

200

200
~ 51.55
- 3.88 lb/min

The decrease in the weight of refrigerant
circulated at the higher suction temperature is

4.15 - 3.88

4.15

x 100

= 6.5%

Since the difference between the vaporizing
and condensing pressures is smaller for the
cycle having the higher suction temperature,
the work of compression per pound required to
compress the vapor from the vaporizing
pressure to the condensing pressure is less for
the higher temperature cycle than for the lower
temperature cycle. It follows then that the heat
of compression per pound for the cycle having
the higher vaporizing temperature is also less
than that for the cycle having the lower vaporiz-
ing temperature. The heat of compression per
pound for the 10° F cycle is 11.54 Btu, whereas
the heat of compression for the 40° F cycle is
only 7.49 Btu. This represents a decrease in the
heat of compression per pound of

(A„ - h e ) - {h d . - h<)

x 100

Because both the work of compression per
pound and the weight of refrigerant circulated

h d — h e

11.54 -7.49

11.54

-35.1%

per minute per ton are less at the higher suction
temperature, the work of compression per
ton and therefore the theoretical horsepower
required per ton will be smaller at the higher
suction temperature. The theoretical horse-
power required per ton of refrigerating capacity
for the 10° F cycle is

42.42

4.15 x (90.90 - 79.36)

42.42

= 1.13

For the 40° F cycle, the theoretical horsepower
required per ton is

"Kb* — hf)
42.42

3.88 x (90.20 - 82.71)

42.42

0.683

In this instance, increasing the vaporizing
temperature of the cycle from 10° F to 40° F
reduces the theoretical horsepower required per
ton by

1.13 -0.683

1.13 X 10 °

= 39.5%

Later, when the efficiency of the compressor
is taken into consideration, it will be shown that
the difference in the actual horsepower required
per ton at the various suction temperatures is
even greater than that indicated by theoretical
computations.

Since the coefficient of performance is an
index of the power required per unit of refriger-
ating capacity and, as such, is an indication of
cycle efficiency, the relative efficiency of the two
cycles can be determined by comparing their
coefficients of performance. The coefficient of
performance for the 10° F cycle is

h c — h a

48.20
= 11.54

= 4.17

100 PRINCIPLES OF REFRIGERATION

and the coefficient of performance for the 40° F
cycle is

hf — h a

h d > - h(f
51.55
~ 7.49
= 6.88

It is evident that the coefficient of perform-
ance, and hence the efficiency of the cycle,
improves considerably as the vaporizing tem-
perature increases. In this instance, increasing
the suction temperature from 10° F to 40° F
increases the efficiency of the cycle by

6.88 - 4.17
4.17

x 100 - 65%

Although the difference in the weight of
refrigerant which must be circulated per minute
per ton of refrigerating capacity at the various
suction temperatures is usually relatively small,
the volume of vapor that the compressor must
handle per minute per ton varies greatly with
changes in the suction temperature. This is
probably one of the most important factors
influencing the capacity and efficiency of a
vapor-compression refrigerating system and is
the one which is the most likely to be overlooked
by the student when studying cycle diagrams.
The difference in the volume of vapor to be
displaced per minute per ton at the various
suction temperatures can be clearly illustrated
by a comparison of the two cycles in question.

For the 10° F cycle, the volume of vapor to be
displaced per minute per ton is

m(v) =4.15 x 1.351 = 5.6cuft

whereas, at the 40° F suction temperature, the
volume of vapor to be displaced per minute per
ton is

ntv) = 3.88 x 0.792 - 3.075 cu ft

It is of interest to note that, whereas the
decrease in the weight of refrigerant circulated
per minute per ton at the higher suction tempera-
ture is only 6.5%, the decrease in the volume
of vapor handled by the compressor per minute
per ton is

5.6 - 3.075

~^— x 100 =45%

Obviously, then, the smaller weight of
refrigerant circulated per minute per ton
accounts for only a very small part of the
reduction in the volume of vapor displaced per
minute per ton at the higher suction tempera-
ture. To a far greater extent, the decrease in
the volume of vapor displaced per minute per
ton is a result of the lower specific volume of the
suction vapor which is coincident with the
higher suction temperature (0.792 cu ft/lb at
40° F as compared to 1.351 cu ft/lb at 10° F).
This aspect of system capacity and efficiency in
relation to suction temperature will be further
investigated in conjunction with compressor
performance in Chapter 12.

The quantity of heat to be rejected at the
condenser per minute per ton is much smaller
for the cycle having the higher suction tempera-
ture. This is true even though the quantity of
heat rejected at the condenser per pound of
refrigerant circulated is nearly the same for
both cycles. For the 10° F cycle, the quantity
of heat rejected at the condenser per minute per
ton is

Mh a -h tt ) = 4.15 x 59.74 = 247.92

whereas for the 40° F cycle the heat rejected at
the condenser per minute per ton is only

"(Ad- - A„) = 3.88 x 59.04 = 229.08 Btu
The quantity of heat rejected per minute per
ton at the condenser is less for the higher
suction temperature because of (1) the smaller
weight of refrigerant circulated per minute per
ton and (2) the smaller heat of compression per
pound.

It has been shown previously that the heat
rejected at the condenser per pound of refriger-
ant circulated is the sum of the heat absorbed
in the evaporator per pound (refrigerating
effect) and the heat of compression per pound.
Since increasing the vaporizing temperature of
the cycle brings about an increase in the
refrigerating effect as well as a decrease in the
heat of compression, the quantity of heat
rejected at the condenser per pound remains
very nearly the same for both cycles (59.74 at
10° F as compared to 59.04 at 40° F). In
general, this is true for all suction temperatures
because any increase or decrease in the heat
of compression is usually accompanied by an
offsetting increase or decrease in the refrigerat-
ing effect.

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE

101

171.8
131.6

29.35

x/**

E'J D'

a/\iw

Eh Z)/^137.5'F

/X !

//l 1 '

/lO*

B 1

// ' i '
fc l! ]!

r i
I
I

/ i 1 ! 1

/i |S? I
/ i i I i

IS
CO

/tol cyl iir> ' |o
/mi id. 1— | 'cm

/ °> Sfi' Is 1 l< T >

/ i II 1 1

A a h a ' h c h e h e ' hi hi'

Specific enthalpy (Btu/lb above - 40° F)

Fig. 7-7. Comparison of two simple saturated cycles operating at different condensing temperatures (figure
distorted). (Refrigerant- 1 2.)

7-1 1. Effect of Condensing Temperature on
Cycle Efficiency. Although the variations in
cycle efficiency with changes in the condensing
temperature are not as great as those brought
about by changes in the vaporizing temperature,
they are nonetheless important. In general,
if the vaporizing temperature remains constant,
the efficiency of the cycle decreases as the
condensing temperature increases, and increases
as the condensing temperature decreases.

To illustrate the effect of condensing tempera-
ture on cycle efficiency, cycle diagrams of two
saturated cycles operating at different condens-
ing temperatures are drawn on the Ph chart in
Fig. 7-7. One cycle, A, B, C, D, and E, has a
condensing temperature of 100° F, whereas the
other cycle, A', B', C, D', and E', is operating at
a condensing temperature of 120° F. The evapo-
rating temperature of both cycles is 10° F. Values
for cycle A-B-C-D-E have been determined in
the previous section. Values for cycle A'-B'-C-
D'-E' are as follows:

From the Ph diagram,

9! = h e - h a - = 79.36 - 36.16 = 43.20 Btu/lb
qi = h d . - h e = 93.20 - 79.36 = 13.84 Btu/lb
?s - h d . - h a . = 93.20 - 36.16 = 57.04 Btu/lb

In a simple saturated cycle the liquid
refrigerant reaches the refrigerant control at

the condensing temperature. Therefore, as
the condensing temperature is increased, the
temperature of the liquid approaching the
refrigerant control is increased and the refriger-
ating effect per pound is reduced. In this
instance, the refrigerating effect is reduced from
48.20 Btu/lb to 43.20 Btu/lb when the condens-
ing temperature is increased from 100° F to
120° F. This is a reduction of

48.20 - 43.20

48.20 X 10 °

- 10.37%

Because the refrigerating effect per pound is
less for the cycle having the higher condensing
temperature, the weight of refrigerant to be
circulated per minute per ton must be greater.
For the cycle having the 100° F condensing
temperature the weight of refrigerant to be
circulated per minute per ton is 4.15 lb. When
the condensing temperature is increased to
120° F, the weight of refrigerant which must be
circulated per minute per ton increases to

200
4T20= 4 - 631b

This is an increase of
4.63 - 4.15

4.15

x 100 = 11.57%

102 PRINCIPLES OF REFRIGERATION

Since the weight of refrigerant which must be
circulated per minute per ton is greater at the
higher condensing temperature, it follows that
the volume of vapor to be compressed per
minute per ton must also be greater. In a simple
saturated cycle the specific volume of the
suction vapor varies only with the vaporizing
temperature. As the vaporizing temperature
is the same for both cycles, the specific volume
of the vapor leaving the evaporator is also the
same for both cycles and therefore the difference
in the volume of vapor to be compressed per
minute per ton is in direct proportion to the
difference in the weight of refrigerant circulated
per minute per ton. At the 100° F condensing
temperature the volume of vapor to be com-
pressed per minute per ton is 5.6 cu ft, whereas
at the 120° F condensing temperature the
volume of vapor compressed per minute per ton
increases to

4.63 x 1.351 =6.25cuft

This represents an increase in the volume of
vapor compressed per minute per ton of

6.25 - 5.6

— x 100 = 11.57%

5.6

Note that the percent increase in the volume
of vapor handled by the compressor is exactly
equal to the percent increase in the weight of
refrigerant circulated. Contrast this with what
occurs when the vaporizing temperature is
varied.

Since the difference between the vaporizing
and condensing pressures is greater, the work
of compression per pound of refrigerant
circulated required to raise the pressure of the
vapor from the vaporizing to the condensing
pressure is also greater for the cycle having the
higher condensing temperature. In this instance,
the heat of compression increases from 11.54
Btu/lb for the 100° F condensing temperature
to 13.84 Btu/lb for the 120° F condensing
temperature. This is an increase of

13.84 - 11.54

— iL54— Xl0 °= 20%
As a result of the greater work of compression
per pound and the greater weight of refrigerant
circulated per minute per ton, the theoretical
horsepower required per ton of refrigerating
capacity increases as the condensing tempera-
ture increases. Whereas the theoretical horse-

power required per ton at the 100° F condensing
temperature is only 1 . 1 3 hp when the condensing
temperature is increased to 120° F, the theoreti-
cal horsepower per ton increases to

4.63 x 13.84

42.42 =L52h P
This is an increase in the power required per ton

Note that the increase in the horsepower
required per ton at the higher condensing
temperature is greater than the increase in the
work of compression per pound. This is
accounted for by the fact that, in addition to the
20% increase in the work of compression per
pound, there is also a 6.5% increase in the
weight of refrigerant circulated per minute per
ton.

The coefficient of performance of the cycle
at the 100° F condensing temperature is 4.17.
When the condensing temperature is raised to
120° F, the coefficient of performance drops to

43.20

= 3.12

13.84

Since the coefficient of performance is an
index of the refrigerating capacity per unit of
power, the decrease in refrigerating capacity per
unit of power in this instance is

4.17 - 3.12

— x 100 = 33.7%

Obviously, the effect of raising the condensing
temperature on cycle efficiency is the exact
opposite of that of raising the evaporating
temperature. Whereas raising the evaporating
temperature increases the refrigerating effect per
pound and reduces the work of compression so
that the refrigerating capacity per unit of power
increases, raising the condensing temperature
reduces the refrigerating effect per pound and
increases the work of compression so that the
refrigerating capacity per unit of power de-
creases.

Although the quantity of heat rejected at the
condenser per pound of refrigerant circulated
varies only slightly with changes in the condens-
ing temperature because any change in the heat
of compression is accompanied by an offsetting
change in the refrigerating effect per pound,

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE

103

Fig. 7-8. Temperature-entropy i

diagram of simple saturated £

cycle on skeleton Ts chart M T *

(figure distorted). (Refrigerant- J

12.) I

the total heat rejected at the condenser per
minute per ton varies considerably with changes
in the condensing temperature because of the
difference in the weight of refrigerant circulated
per minute per ton. It was shown in Section 7-7
that the total heat rejected at the condenser per
minute per ton (gj) is always the sum of the
heat absorbed in the evaporator per minute per
ton (Q x ) and the total heat of compression per
ton (Q 2 ). Since Q x is always 200 Btu/min/ton,
then Q 3 will vary only with Q 2 , the heat of
compression per minute per ton. Furthermore,
since Q 2 always increases as the condensing
temperature increases, it follows then that Q 3
also increases as .the condensing temperature
increases.

For the two cycles in question, the heat
rejected at the condenser per minute per ton
at the 100° F condensing temperature is 218.75
Btu. For the 120° F condensing temperature,
the quantity of heat rejected at the condenser per
minute per ton increases to

4.63 x (43.20 + 13.84) = 310.4 Btu
The percent increase is
310.40 - 218.75
— lii™ >< 100 =41-8%

It is interesting to note also that the amount
of sensible heat rejected at the condenser in-
creases considerably at the higher condensing
temperature, whereas the amount of latent heat
rejected diminishes slightly. This indicates that,
at the higher condensing temperature, a

Specific entropy (Btu/lb *F)

greater portion of the condenser surface is being
used merely to reduce the temperature of the
discharge vapor to the condensing temperature.
7-12. The Temperature-Entropy Diagram.
Although the author is partial to the pressure-
enthalpy diagram, there are many who prefer
to use the temperature-entropy diagram to
analyze the refrigeration cycle. To acquaint
the student with the use of Ts diagrams in'cycle
analysis, a diagram of the simple saturated cycle
described in the foregoing sections is drawn
on Ts coordinates in Fig. 7-8. The state points
A, B, C, D, and E represent the points in the
cycle as shown by the flow diagram in Fig. 7-5.
The state point X represents saturated liquid
at the vaporizing temperature. Process A-B is
the irreversible adiabatic expansion through the
refrigerant control.* Process B-C is the iso-
baric and isothermal vaporization in the
evaporator. Process C-D is the reversible
adiabatic (isentropic) compression in the com-
pressor and processes D-E and E-A are the
desuperheating and condensing processes in the
condenser. T t , T c , and T d are the absolute
suction, absolute condensing, and absolute
discharge temperatures, respectively, whereas
s a , s b , s e , s d , s t , and s x are the specific entropies
of the refrigerant at the various state points.

The principal advantage of the Ts diagram is
that the areas shown on the chart represent
actual heat quantities. In Fig. 7-8, the area

* The line A-B does not necessarily follow the
actual path of process A-B. See Section 7-4.

104 PRINCIPLES OF REFRIGERATION

Condensing Temperature, 100° F

Condensing Pressure, 136.16 Psia

Suction temperature

50°

40°

30°

20°

10°

0°

-10°

-20°

-30°

-40°

Absolute suction pressure

61.39

51.68

43.16

35.75

29.35

23.87

19.20

15.28

12.02

9.32

Refrigerating effect per
pound

52.62

51.55

50.45

49.33

48.20

47.05

45.89

44.71

43.54

42.34

Weight of refrigerant
circulated per minute per
ton

3.80

3.88

3.97

4.05

4.15

4.25

4.36

4.48

4.59

4.73

Specific volume of suction
vapor

0.673

0.792

0.939

1.121

1.351

1.637

2.00

2.47

3.09

3.91

Volume of vapor com-
pressed per minute per ton

2.56

3.08

3.77

4.55

5.60

6.96

8.72

11.10

14.20

18.50

Heat of compression per
pound

6.01

7.49

8.79

10.11

11.54

13.29

14.85

16.73

18.50

20.40

Theoretical horsepower per
ton

0.539

0.683

0.818

0.965

1.13

1.35

1.54

1.78

2.00

2.26

Coefficient of performance

8.76

6.88

5.74

4.88

4.17

3.54

3.09

2.67

2.36

2.07

Fig. 7-9

^40 -30 -20 -10 10 20 30 40 50
Suction temperature

4
3.5

3.0 a

2.5 f
2-0 1

1.5
1.0
0.5

20
18
16

14 c
12 -
10 |
8 *
6
4
2

Fig. 7-IOa. For Refrigerant- 1 2, the refrigerating effect per pound, the weight of refrigerant circulated per
minute per ton, the specific volume of the suction vapor, and the volume of vapor compressed per minute
per ton are each plotted against suction temperature. Condensing temperature is constant at 100° F.

CYCLE DIAGRAMS AND THE SIMPLE SATURATED CYCLE 105

Fig. 7-IOb. For Refrigerant- 1 2,
the coefficient of performance
and the horsepower per ton are
plotted against suction tem-
perature. Condensing tempera-
ture is constant at 100° F.

2.6
2.2

£
o

1.6

1.2

0.8

0.4

cop-P"*

^Horserjower

per ton

-40 -

A-X-C-D-E-A represents the heat energy
equivalent of the work of adiabatic compression
and, since the distance between the base line and
T„ foreshortened in the figure, represents the
absolute vaporizing temperature of the liquid,
area B-C-Sc-sg-B represents the refrigerating
effect per pound. The sum of the areas A-X-C-
D-E-A and B-C-Sg-sg-B, of course, represents
the heat rejected at the condenser per pound.
As in the case of the Ph diagram, it can be
readily seen on the Ts chart that either lowering
the vaporizing temperature or raising the con-
densing temperature tends to increase the work
of compression, reduce the refrigerating effect
per pound, and lower the efficiency of the cycle.
7-13. Summary. Regardless of the method
used to analyze the cycle, it is evident that the
capacity and efficiency of a refrigerating system
improve as the vaporizing temperature increases
and as the condensing temperature decreases.
Obviously, then, a refrigerating system should
always be designed to operate at the highest
possible vaporizing temperature and the lowest
possible condensing temperature commensurate
with the requirements of the application. This
will nearly always permit the most effective use
of the smallest possible equipment and thereby
effect a savings not only in the initial cost
of the equipment but also in the operating
expenses.

In any event, the influence of the vaporizing
and condensing temperatures on cycle efficiency
is of sufficient importance to warrant a more
intensive study. To aid the student in this, the
relationship between the refrigerating effect per

8 g

■£
5 S

30-20-10 10 20 30 40 50
Suction temperature

pound, the weight of refrigerant circulated per
minute per ton, the specific volume of the suction
vapor, the volume of vapor compressed per
minute per ton, the horsepower required per
ton, and the coefficient of performance of the
cycle has been calculated for various suction
temperatures. These values are given in tabular
form in Fig. 7-9 and are illustrated graphically
in Figs. 7-10a and 7-106. In addition, the effect
of condensing temperature on the horsepower
required per ton of refrigerating capacity is
shown for several condensing temperatures in
Fig. 7-11.

Since the properties of the refrigerant at point
D on the cycle diagram cannot ordinarily be
obtained from the refrigerant tables and since
these properties are difficult to read accurately
from the Ph chart because of the size of the
chart, the approximate isentropic discharge

-40 -30 -20 -10 10 20 30 40 50
Suction temperature

Fig. 7-11. The effect of condensing temperature on
the horsepower per ton.

v i

*r1

H„

fcfc?

*£

■>o

fc^l

-^_

^"s;

^jT

— ~.

106 PRINCIPLES OF REFRIGERATION

temperatures and the approximate enthalpy
of the refrigerant vapor at point D have been
compiled for a variety of vaporizing and con-
densing temperatures and are given in Table 7-1
to aid the student in arriving at more accurate
solutions to the problems at the end of the
chapter.

PROBLEMS

1. A Refrigerant-12 system operating on a
simple saturated cycle has an evaporating tem-
perature of 0° F and a condensing temperature
of 110° F. Determine:
A. (1) The refrigerating effect per pound of
refrigerant circulated.

Arts. 44.56 Btu/lb
(2) The weight of refrigerant circulated
per minute per ton.

Arts. 4.49 lb/min/ton

(3) The volume of vapor compressed per
minute per ton.

Ans. 7.35 cu ft/min/ton

B. (1) The heat of compression per pound of

refrigerant circulated.

Ans. 14.39 Btu/lb

(2) The heat of compression per minute
per ton of refrigeration.

Ans. 64.61 Btu/min/ton

(3) The work of compression per minute
per ton in foot-pounds.

Ans. 50.267 lb/min/ton

(4) The theoretical horsepower per ton.

Ans. 1.52hp/ton

(5) The coefficient of performance.

Ans. 3.1

C. (1) The heat rejected per minute per ton

at the condenser.

Ans. 264.61 Btu/min/ton

Actual

Refrigerating

Cycles

8-1. Deviation from the Simple Saturated
Cycle. Actual refrigerating cycles deviate some-
what from the simple saturated cycle. The reason
for this is that certain assumptions are made for
the simple saturated cycle which do not hold
true for actual cycles. For example, in the
simple saturated cycle, the drop in pressure in
the lines and across the evaporator, condenser,
etc., resulting from the flow of the refrigerant
through these parts is neglected. Furthermore,
the effects of subcooling the liquid and of super-
heating the suction vapor are not considered.
Too, compression in the compressor is assumed
to be true isentropic compression. In the
following sections all these things are taken into
account and their effect on the cycle is studied
in detail.*

8-2. The Effect of Superheating the Suction
Vapor. In the simple saturated cycle, the suction
vapor is assumed to reach the suction inlet of
the compressor as a saturated vapor at the
vaporizing temperature and pressure. In actual
practice, this is rarely true. After the liquid
refrigerant has completely vaporized in the
evaporator, the cold, saturated vapor will
usually continue to absorb heat and thereby
become superheated before it reaches the
compressor (Fig. 8-1).

* The departure from true isentropic compression
and the effect that it has on the cycle are discussed
in Chapter 12.

On the Ph diagram in Fig. 8-2, a simple
saturated cycle is compared to one in which the
suction vapor is superheated from 20° F to
70° F. Points A , B, C, D, and E mark the satur-
ated cycle, and points A, B, C", D', and E indi-
cate the superheated cycle.

If the slight pressure drop resulting from the
flow of the vapor in the suction piping is
neglected, it may be assumed that the pressure
of the suction vapor remains constant during
the superheating. That is, after the superheat-
ing, the pressure of the vapor at the suction
inlet of the compressor is still the same as
the vaporizing pressure in the evaporator.
With this assumption, point C'can be located on
the Ph chart by following a line of constant
pressure from point C to the point where the
line of constant pressure intersects the 70° F
constant temperature line. Point D' is found by
following a line of constant entropy from point
C to the line of constant pressure corresponding
to the condensing pressure.

In Fig. 8-2, the properties of the superheated
vapor at points C" and D', as read from the Ph
chart, are as follows:

At point C",

p = 35.75 psia / = 70° F

v = 1 .260 cu ft/lb h = 88.6 Btu/lb
s = 0.1840 Btu/lb/° R

, 20*F

35.75 psia

-*&■

20* F

35.75 psia

30"F C 20* F

' 35.75 psia 4 \

50* F

70*F

Saturated vapor
■Superheated vapor

^ XT 164* F_

35.75 r^i 131.6 psi'a
psia

100° E

100* F

131.6 psia ")

A 100* F

/^\

131.6 psia

Fig. 8-1. Flow diagram of superheated cycle. Liquid
completely vaporized at point C — saturated vapor
continues to absorb heat while flowing from C to
C — vapor reaches compressor in superheated
condition. Notice the high discharge temperature.
(Refrigerant- 1 2.)

107

108 PRINCIPLES OF REFRIGERATION

131.16

£ 35.75

Superheat

Enthalpy (Btu/lb)
Fig. 8-2. Ph diagram comparing simple saturated cycle to the superheated cycle. (Refrigerant- 1 2).

At point/)',
/> = 131.6 psia f = 164°F

v - 0.380 cu ft/lb h = 99.2 Btu/lb
j = 0.1840 Btu/lb/° R

On the Ph chart, process C-C represents the
superheating of the suction vapor from 20° F
to 70° F at the vaporizing pressure, and the
difference between the enthalpy of the vapor
at these points is the amount of heat required to
superheat each pound of refrigerant. In
comparing the two cycles, the following
observations are of interest:

1 . The heat of compression per pound for the
superheated cycle is slightly greater than that
for the saturated cycle. For the superheated
cycle, the heat of compression is

h x - h & = 99.2 - 88.6 = 10.6 Btu/lb

whereas for the saturated cycle the heat of
compression is

h d -h e = 90.6 - 80.49 = 10.11 Btu/lb

In this instance, the heat of compression per
pound is greater for the superheated cycle by

10.6 - 10.11

10.11

x 100 =» 4.84%

2. For the same condensing temperature and
pressure, the temperature of the discharge vapor
leaving the head of the compressor is consider-

ably higher for the superheated cycle than for
the saturated cycle — in this case, 164° F for the
superheated cycle as compared to 1 12° F for the
saturated cycle.

3. For the superheated cycle, a greater
quantity of heat must be dissipated at the
condenser per pound than for the saturated
cycle. This is because of the additional heat
absorbed by the vapor in becoming superheated
and because of the small increase in the heat of
compression per pound. For the superheated
cycle, the heat dissipated at the condenser per
pound is

h- ~ K - 99.2 - 31.16 = 68.04 Btu/lb

and for the saturated cycle the heat dissipated
at the condenser per pound is

h d -h a = 90.6 - 31.16 = 59.44 Btu/lb

The percent increase in the heat dissipated at

the condenser per pound for the superheated

cycle is

68.04-59.44 tnn ijjd ,
— x 100 = 14.4%

Note that the additional heat which must be
dissipated per pound at the condenser in the
superheated cycle is all sensible heat. The
amount of latent heat dissipated per pound is
the same for both cycles. This means that in the
superheated cycle a greater amount of sensible
heat must be given up to the condensing medium

ACTUAL REFRIGERATING CYCLES

109

before condensation begins and that a greater
portion of the condenser will be used in cooling
the discharge vapor to its saturation temperature.
Notice also that, since the pressure of the
suction vapor remains constant during the
superheating, the volume of the vapor increases
with the temperature approximately in accord-
ance with Charles' law.* Therefore, a pound
of superheated vapor will always occupy a
greater volume than a pound of saturated vapor
at the same pressure. For example, in Fig. 8-2,
the specific volume of the suction vapor increases
from 1.121 cu ft per pound at saturation to
1 .260 cu ft per pound when superheated to 70° F.
This means that for each pound of refrigerant
circulated, the compressor must compress a
greater volume of vapor if the vapor is super-
heated than if the vapor is saturated. For this
reason, in every instance where the vapor is
allowed to become superheated before it reaches
the compressor, the weight of refrigerant cir-
culated by a compressor of any given displace-
ment will always be less than when the suction
vapor reaches the compressor in a saturated
condition, provided the pressure is the same.

The effect that superheating of the suction
vapor has on the capacity of the system and on
the coefficient of performance depends entirely
upon where and how the superheating of the
vapor occurs and upon whether or not the
heat absorbed by the vapor in becoming super-
heated produces useful cooling.*
8-3. Superheating without Useful Cooling.
Assume first that the superheating of the
suction vapor occurs in such a way that no use-
ful cooling results. When this is true, the refri-
gerating effect per pound of refrigerant circulated
is the same for the superheated cycle as for a
saturated cycle operating at the same vaporizing
and condensing temperatures, and therefore the
weight of refrigerant circulated per minute per
ton will also be the same for both the superheated
and saturated cycles. Then, for both cycles
illustrated in Fig. 8-2,

* The temperature and volume of the vapor do
not vary exactly in accordance with Charles' law
because the refrigerant vapor is not a perfect gas.

* The effects of superheating depend also upon
the refrigerant used. The discussion in this chapter
is limited to systems using R-12. The effects of
superheating on systems using other refrigerants are
discussed later.

The weight of refrig-

200

erant circulated per

fie ~ K

minute per ton m

200

49.33

= 4.05 lb/min/ton

Since the weight of refrigerant circulated is the
same for both the superheated and saturated
cycles and since the specific volume of the vapor
at the compressor inlet is greater for the super-
heated cycle than for the saturated cycle, it
follows that the volume of vapor that the com-
pressor must handle per minute per ton of
refrigerating capacity is greater for the super-
heated cycle than for the saturated cycle.

For the saturated
cycle, the specific
volume of the suction
vapor v e

The volume of
vapor compressed per
minute per ton V

For the superheated
cycle, the specific
volume of the suction
vapor «v

The volume of vapor
compressed per minute
per ton V

= 1.121 cu ft/lb

= m x v

= 4.05 x 1.121

= 4.55 cu ft/min/ton

= 1.260 cu ft/lb

= m x v

= 4.05 x 1.260

= 5.02 cu ft/min/ton

In regard to percentage, the increase in the
volume of vapor which must be handled by a
compressor operating on the superheated cycle

This means, of course, that a compressor
operating on the superheated cycle must be
10.3% larger than the one required for the
saturated cycle.

Again, since the weight of refrigerant circu-
lated per minute per ton is the same for both
cycles and since the heat of compression per
pound is greater for the superheated cycle than
for the saturated cycle, it is evident that the
horsepower per ton is greater for the super-
heated cycle and the coefficient of performance
is less.

110 PRINCIPLES OF REFRIGERATION

For the saturated cycle,

m(h d - h e )

the horsepower per ton

42.42

4.05 x 10.11

42.42

= 0.965 hp/ton

The coefficient of per-

h e — h a

formance

h d -K

49.33

10.11

= 4.88

For the superheated

m(h d - - h C ')

cycle, the horsepower per

42.42

ton

4.05 x 10.6

42.42

= 1.01 hp/ton

The coefficient of per-

h c ~h a

formance

h d ' - K-

49.33

10.60

= 4.65

In summary, when superheating of the vapor
occurs without producing useful cooling, the
volume of vapor compressed per minute per ton,
the horsepower per ton, and the quantity of heat
given up in the condenser per minute per ton are
all greater for the superheated cycle than for the
saturated cycle. This means that the com-
pressor, the compressor driver, and the conden-
ser must all be larger for the superheated cycle
than for the saturated cycle.
8-4. Superheating That Produces Useful
Cooling. Assume, now, that all of the heat
taken in by the suction vapor produces useful
cooling. When this is true, the refrigerating effect
per pound is greater by an amount equal to the
amount of superheat. In Fig. 8-2, assuming that
the superheating produces useful cooling, the
refrigerating effect per pound for the superheated
cycle is equal to

h^ -h a = 88.60 - 31.16 = 57.44 Btu/lb
Since the refrigerating effect per pound is
greater for the superheated cycle than for the
saturated cycle, the weight of refrigerant circu-
lated per minute per ton is less for the super-
heated cycle than for the saturated cycle.
Whereas the weight of refrigerant circulated per
minute per ton for the saturated cycle is 4.05, the

weight of refrigerant circulated per minute per
ton for the superheated cycle is

200 200

= 3.48 lb/min/ton

h- - h n

57.44

Notice that, even though the specific volume
of the suction vapor and the heat of compression
per pound are both greater for the superheated
than for the saturated cycle, the volume of vapor
compressed per minute per ton and the horse-
power per ton are less for the superheated cycle
than for the saturated cycle. This is because of
the reduction in the weight of refrigerant circu-
lated. The volume of vapor compressed per
minute per ton and the horsepower per ton for
the saturated cycle are 4.55 cu ft and 0.965 hp,
respectively, whereas for the superheated cycle

The volume of = m x v c -

vapor compressed per =3.48 x 1.260
minute per ton V = 4.38 cu ft/min/ton

The horsepower per m{h a > — A c .)

ton 42.42

3.48 x 10.60

42.42
= 0.870 hp/ton

For the superheated cycle, both the refrig-
erating effect per pound and the heat of com-
pression per pound are greater than for the
saturated cycle. However, since the increase in
the refrigerating effect is greater proportionally
than the increase in the heat of compression, the
coefficient of performance for the superheated
cycle is higher than that of the saturated cycle.
For the saturated cycle, the coefficient of per-
formance is 4.69, whereas for the superheated
cycle

The coefficient
of performance

(h c . - h a ) 57.44

= 5.42

(4r - h c ) 10.60
It will be shown in the following sections that
the superheating of the suction vapor in an
actual cycle usually occurs in such a way that a
part of the heat taken in by the vapor in becom-
ing superheated is absorbed from the refrig-
erated space and produces useful cooling,
whereas another part is absorbed by the vapor
after the vapor leaves the refrigerated space and
therefore produces no useful cooling. The por-
tion of the superheat which produces useful
cooling will depend upon the individual applica-
tion, and the effect of the superheating on the

ACTUAL REFRIGERATING CYCLES

cycle will vary approximately in proportion to
the useful cooling accomplished.

Regardless of the effect on capacity, except
in some few special cases, a certain amount of
superheating is nearly always necessary and, in
most cases, desirable. When the suction vapor
is drawn directly from the evaporator into the
suction inlet of the compressor without at least
a small amount of superheating, there is a good
possibility that small particles of unvaporized
liquid will be entrained in the vapor. Such a
vapor is called a "wet" vapor. It will be shown
later that "wet" suction vapor drawn into the
cylinder of the compressor adversely affects the
capacity of the compressor. Furthermore, since
refrigeration compressors are designed as vapor
pumps, if any appreciable amount of un-
vaporized liquid is allowed to enter the com-
pressor from the suction line, serious mechanical
damage to the compressor may result. Since
superheating the suction vapor eliminates the
possibility of "wet" suction vapor reaching the
compressor inlet, a certain amount of super-
heating is usually desirable. Again, the extent
to which the suction vapor should be allowed to
become superheated in any particular instance
depends upon where and how the superheating
occurs and upon the refrigerant used.

Superheating of the suction vapor may take
place in any one or in any combination of the
following places:

1. In the end of the evaporator

2. In the suction piping installed inside the
refrigerated space (usually referred to as a "drier
loop")

3. In the suction piping located outside of the
refrigerated space

4. In a liquid-suction heat exchanger.

8-5. Superheating in Suction Piping outside
the Refrigerated Space. When the cool re-
frigerant vapor from the evaporator is allowed
to become superheated while flowing through
suction piping located outside of the refrigerated
space, the heat taken in by the vapor is absorbed
from the surrounding air and no useful cooling
results. It has already been demonstrated that
superheating of the suction vapor which pro-
duces no useful cooling adversely affects the
efficiency of the cycle. Obviously, then, super-
heating of the vapor in the suction line outside

of the refrigerated space should be eliminated
whenever practical.

Superheating of the suction vapor in the
suction line can be prevented for the most part
by insulating the suction line. Whether or not
the loss of cycle efficiency in any particular
application is sufficient to warrant the additional
expense of insulating the suction line depends
primarily on the size of the system and on the
operating suction temperature.

When the suction temperature is relatively
high (35° F or 40° F), the amount of superheating
will usually be small and the effect on the
efficiency of the cycle will be negligible. The
reverse is true, however, when the suction tem-
perature is low. The amount of superheating is
apt to be quite large.

Too, at low suction temperatures, when the
efficiency of the cycle is already very low, each
degree of superheat will cause a greater reduc-
tion in cycle efficiency percentagewise than when
the suction temperature is high. It becomes
immediately apparent that any appreciable
amount of superheating in the suction line of
systems operating at low suction temperatures
will seriously reduce the efficiency of the cycle
and that, under these conditions, insulating of
the suction line is not only desirable but
absolutely necessary if the efficiency of the cycle
is to be maintained at a reasonable level.

Aside from any considerations of capacity,
even at the higher suction temperatures, insulat-
ing of the suction line is often required to prevent
frosting or sweating of the suction line. In
flowing through the suction piping, the cold
suction vapor will usually lower the temperature
of the piping below the dew point temperature
of the surrounding air so that moisture will
condense out of the air onto the surface of the
piping, causing the suction piping to either frost
or sweat, depending upon whether or not the
temperature of the piping is below the freezing
temperature of water. In any event, frosting or
sweating of the suction piping is usually un-
desirable and should be eliminated by insulating
the piping.

8-6. Superheating the Vapor inside the Re-
frigerated Space. Superheating of the suction
vapor inside the refrigerated space can take
place either in the end of the evaporator or in
suction piping located inside the refrigerated
space, or both.

112 PRINCIPLES OF REFRIGERATION

Fig. 8-3. Flow diagram showing drier loop for super-
heating suction vapor inside refrigerated space.

To assure the proper operation of the refri-
gerant control and to prevent liquid refrigerant
from overflowing the evaporator and being
carried back to the compressor, when certain
types of refrigerant controls are used, it is
necessary to adjust the control so that the liquid
is completely evaporated before it reaches the
end of the evaporator. In such cases, the cold
vapor will continue to absorb heat and become
superheated as it flows through the latter portion
of the evaporator. Since the heat to superheat
the vapor is drawn from the refrigerated space,
useful cooling results and the refrigerating effect
of each pound of refrigerant is increased by an
amount equal to the amount of heat absorbed in
the superheating.

It has been shown that when the superheating
of the suction vapor produces useful cooling the
efficiency of the cycle is improved somewhat.*
However, in spite of the increase in cycle effi-
ciency, it must be emphasized that superheating
the suction vapor in the evaporator is not
economical and should always be limited to only
that amount which is necessary to the proper
operation of the refrigerant control. Since the
transfer of heat through the walls of the evapora-
tor per degree of temperature difference is not as

* Although this is true for systems using R-12 as
a refrigerant, it will be shown later that this is not
true for all refrigerants.

great to a vapor as to a liquid, the capacity of the
evaporator is always reduced in any portion of
the evaporator where only vapor exists. There-
fore, excessive superheating of the suction vapor
in the evaporator will reduce the capacity of the
evaporator unnecessarily and will require either
that the evaporator be operated at a lower
vaporizing temperature or that a larger evapora-
tor be used in order to provide the desired
evaporator capacity. Neither of these is desir-
able nor practical. Since the space available for
evaporator installation is often limited and since
evaporator surface is expensive, the use of a
larger evaporator is not practical. Because of
the effect on cycle efficiency, the undesirability of
lowering the vaporizing temperature is obvious.
Often, a certain amount of suction piping,
usually called a drier loop, is installed inside the
refrigerated space for the express purpose of
superheating the suction vapor (Fig. 8-3). Use
of a drier loop permits more complete flooding
of the evaporator with liquid refrigerant without
the danger of the liquid overflowing into the
suction line and being drawn into the compres-
sor. This not only provides a means of super-
heating the suction vapor inside the refrigerated
space so that the efficiency of the cycle is
increased without the sacrifice of expensive eva-
porator surface, but it actually makes possible
more effective use of the existing evaporator
surface. Also, in some instances, particularly
where the suction temperature is high and the
relative humidity of the outside air is reasonably
low, superheating of the suction vapor inside the
refrigerated space will raise the temperature of
the suction piping and prevent the formation of
moisture, thereby eliminating the need for suction
line insulation. It should be noted, however,
that the extent to which the suction vapor can be
superheated inside the refrigerated space is
limited by the space temperature. Ordinarily, if
sufficient piping is used, the suction vapor can
be heated to within 4° F to 5° F of the space
temperature. Thus, for a 40° F space tempera-
ture, the suction vapor may be superheated to
approximately 35° F.

8-7. The Effects of Subcooling the Liquid.
On the PA diagram in Fig. 8-4, a simple saturated
cycle is compared to one in which the liquid is
subcooled from 100° F to 80° F before it reaches
the refrigerant control. Points A, B, C, D, and E
designate the simple saturated cycle, whereas

points A', ff, C, D, and E designate the sub-
cooled cycle.

It has been shown (Section 6-28) that when
the liquid is subcooled before it reaches the
refrigerant control the refrigerating effect per
pound is increased. In Fig. 8-4, the increase in
the refrigerating effect per pound resulting from
the subcooling is the difference between h h and
h v , and is exactly equal to the difference between
h a and h a ; which represents the heat removed
from the liquid per pound during the subcooling.

For the saturated cycle,
the refrigerating effect per
pound, q x

For the subcooled cycle,
the refrigerating effect per
pound, q t

— K — K

= 80.49 -31.16
= 49.33 Btu/lb
= h e — h tt >
= 80.49 - 26.28
= 54.21 Btu/lb

200
~ 49.33
= 4.05 lb

200

Because of the greater refrigerating effect per
pound, the weight of refrigerant circulated per
minute per ton is less for the subcooled cycle
than for the saturated cycle.

For the saturated cycle, the
weight of refrigerant circulated
per minute per ton m

For the subcooled cycle, the
weight of refrigerant circulated — 54.21
per minute per ton m = 3.69 lb

Notice that the condition of the refrigerant
vapor entering the suction inlet of the compressor
is the same for both cycles. For this reason, the
specific volume of the vapor entering the com-
pressor will be the same for both the saturated
and subcooled cycles and, since the weight of
refrigerant circulated per minute per ton is less

ACTUAL REFRIGERATING CYCLES 113

for the subcooled cycle than for the saturated
cycle, it follows that the volume of vapor which
the compressor must handle per minute per ton
will also be less for the subcooled cycle than for
the saturated cycle.

For the saturated cycle,
the specific volume of the
suction vapor v e

The volume of vapor
compressed per minute per
ton V

For the subcooled cycle,
the specific volume of the
suction vapor v e

The volume of vapor
compressed per minute per
ton V

1.121 cu ft/lb
m x v e
4.05 x 1.121
■■ 4.55 cu ft/min

■ 1.121 cu ft/lb

m x v e

3.69 x 1.121
: 4.15 cu ft/min

Because the volume of vapor compressed per
minute per ton is less for the subcooled cycle,
the compressor displacement required for the
subcooled cycle is less than that required for the
saturated cycle.

Notice also that the heat of compression per
pound and therefore the work of compression
per pound is the same for both the saturated and
subcooled cycles. This means that the refri-
gerating effect per pound resulting from the
subcooling is accomplished without increasing
the energy input to the compressor. Any change
in the refrigerating cycle which increases the
quantity of heat absorbed in the refrigerated
space without causing an increase in the energy
input to the compressor will increase the c.o.p.
of the cycle and reduce the horsepower required
per ton.

Fig. 8-4. Ph diagrams compar-
ing the subcooled cycle to
the simple saturated cycle.
(Refrigerant- 1 2.)

131.16

: 35.75

Enthalpy (Btu/lb)

114 PRINCIPLES OF REFRIGERATION

Liquid-vapor mixture
35.75 psia-20°F

Superheated vapor
131.16 paa-112'F

Saturated vapor
3575psia-20*F

Saturated vapor
131.16 psia-lOOV

Subcooled liquid
131.16 psia-SCF

Liquid subcooled 20*
by giving off heat to
surrounding air while
passing through
liquid line, receiver,
etc. ,

Fig. 8-5. Flow diagram illus-
trating subcooling of the liquid
in the liquid line. (Refrigerant-
12.)

)

For the saturated cycle,
the coefficient of perfor-
mance

The horsepower per ton

For the subcooled cycle,
the coefficient of perfor-
mance

h d - K
80.49 -

- 26.28

90.60 -
54.21
10.51
= 5.16
m{h d -

• 80.49
h e )

42.42
3.69 x 10.51

Saturated liquid
131.16 psia-100 o F

K - K

hd — h c

80.49 - 31.16
~ 90.60 - 80.49
= 4.88
_ m(h d — h c )

42A2
= 4.05 x 10.51
= 0.965 hp/ton

h c . — h„'

The horsepower per ton

42.42
= 0.914 hp/ton

In this instance, the c.o.p. of the subcooled cycle
is greater than that of the saturated cycle by

5.H5 -4.88

~^M~ X 10 ° = 5 - 7%
Subcooling of the liquid refrigerant can and
does occur in several places and in several ways.
Very often the liquid refrigerant becomes sub-
cooled while stored in the liquid receiver tank
or while passing through the liquid line by giving
off heat to the surrounding air (Fig. 8-5). In
some cases where water is used as the condensing
medium, a special liquid .subcooler is used to
subcool the liquid (Fig. 8-6). The gain in system
capacity and efficiency resulting from the liquid

subcooling is very often more than sufficient to
offset the additional cost of the subcooler,
particularly for low temperature applications.
The liquid subcooler may be piped either in
series or in parallel with the condenser. When
the subcooler is piped in series with the con-
denser, the cooling water passes through the
subcooler first and then through the condenser,
thereby bringing the coldest water into contact
with the liquid being subcooled (Fig. 8-7).
There is some doubt about the value of a sub-
cooler piped in series with the condenser. Since
the cooling water is warmed by the heat absorbed
in the subcooler, it reaches the condenser at a
higher temperature and the condensing tempera-
ture of the cycle is increased . Hence the increase
in system efficiency resulting from the subcooling
is offset to some extent by the rise in the con-
densing temperature.

When the subcooler is piped in parallel with
the condenser (Fig. 8-6), the temperature of the
water reaching the condenser is not affected by
the subcooler. However, for either series or
parallel piping, the size of the condenser water
pump must be increased when a subcooler is
added. If this is not done, the quantity of water
circulated through the condenser will be
diminished by the addition of the subcooler and
the condensing temperature of the cycle will be
increased, thus nullifying any benefit accruing
from the subcooling.

Notice that in each case discussed so far,
the heat given up by the liquid in becoming
subcooled is given up to some medium external
to the system.

8-8. Liquid-Suction Heat Exchangers. An-
other method of subcooling the liquid is to bring
about an exchange of heat between the liquid

and the cold suction vapor going back to the
compressor. In a liquid-suction heat exchanger,
the cold suction vapor is piped through the
heat exchanger in counterflow to the warm
liquid refrigerant flowing through the liquid
line to the refrigerant control (Fig. 8-8). In
flowing through the heat exchanger the cold
suction vapor absorbs heat from the warm
liquid so that the liquid is subcooled as the
vapor is superheated, and, since the heat
absorbed by the vapor in becoming superheated
is drawn from the liquid, the heat of the liquid
is diminished by an amount equal to the amount
of heat taken in by the vapor. In each of the
methods of subcooling discussed thus far, the
heat given up by the liquid in becoming sub-
cooled is given up to some medium external to
the system and the heat then leaves the system.
When a liquid-suction heat exchanger is used,
the heat given up by the liquid in becoming
subcooled is absorbed by the suction vapor and
remains in the system.

On the Ph diagram in Fig. 8-9, a simple
saturated cycle is compared to one in which a
liquid-suction heat exchanger is employed.
Points A, B, C, Z),and £ identify the saturated
cycle and points A', B', C", D', E identify the
in which the heat exchanger is used. In the cycle
latter cycle, it is assumed that the suction vapor
is superheated from 20° F to 60° F in the heat
exchanger.

The heat absorbed per pound of vapor in the
heat exchanger is

h c . -h c = 86.20 - 80.49 = 5.71 Btu/lb

D 5

Water from subcooler

Water-cooled

condenser

(10O*F condensing)

t/80-

100*

75' »

i tower

or city main

Liquid to
'subcooler

90* water to cooling tower or sewer

Fig. 8-4. Flow diagram illustrating subcooler piped
in series with condenser.

ACTUAL REFRIGERATING CYCLES

115

Water from^/
condenser

Saturated liquid
to subcooler

Fig. 8-7. Flow diagram showing parallel piping for
condenser and subcooler.

Since the heat given up by the liquid in
the heat exchanger in becoming subcooled is
exactly equal to the heat absorbed by the vapor
in becoming superheated, h a — h a - is equal to
h C ' — h c and therefore is also equal to 5.71
Btu/lb. Since h a — h a ' represents an increase
in the refrigerating effect, the refrigerating
effect per pound for the heat exchanger cycle is

h c - h a . = 80.49 - 25.45 = 55.04

The heat of compression per pound for the heat
exchanger cycle is

h a . - h d = 97.60 - 86.20 = 11.40

Therefore, the coefficient of performance is

K - h n . 55.04

h,.-h, 11.40

= 4.91

The coefficient of performance of the satur-
ated cycle is 4.88. Therefore, the coefficient of
performance of the heat exchanger cycle is
greater than that of the saturated cycle by only

4.91 -4.88

4.88

x 100 =0.5%

Depending upon the particular case, the
coefficient of performance of a cycle employing
a heat exchanger may be either greater than,
less than, or the same as that of a saturated
cycle operating between the same pressure
limits. In any event, the difference is negligible,
and it is evident that the advantages accruing

116 PRINCIPLES OF REFRIGERATION

20* vaporizing
temperature

Saturated
suctio n f
vapor- v
20'

Subcooled
liquid-75"F~

Saturated
liquid-lOOT

Fig. 8-8. Flow diagram of refrigera-
tion cycle illustrating the use of a
liquid-suction heat exchanger.

100* condensing
temperature

from the subcooling of the liquid in the heat
exchanger are approximately offset by the
disadvantages of superheating the vapor.
Theoretically, then, the use of a heat exchanger
cannot be justified on the basis of an increase in
system capacity and efficiency. However, since
in actual practice a refrigerating system does
not (cannot) operate on a simple saturated
cycle, this does not represent a true appraisal of
the practical value of the heat exchanger.
In an actual cycle, the suction vapor will

131.16

always become superheated before the com-
pression process begins because nothing can be
done to prevent it. This is true even if no
superheating takes place either in the evapora-
tor or in the suction line and the vapor reaches
the inlet of the compressor at the vaporizing
temperature. As the cold suction vapor flows
into the compressor, it will become superheated
by absorbing heat from the hot cylinder walls.
Since the superheating in the compressor cyl-
inder will occur before the compression process

35.75 ,

Enthalpy (Btu/lb)

Hg. 8-». Ph diagrams comparing simple saturated cycle to cycle employing a liquid-suction heat exchanger.
The amount of subcooling is equal to the amount of superheating. (Refrigerant- 1 2.)

begins, the effect of the superheating on cycle
efficiency will be approximately the same as if
the superheating occurred in the suction line
without producing useful cooling.*

The disadvantages resulting from allowing the
suction to become superheated without pro-
ducing useful cooling have already been pointed
out. Obviously, then, since superheating of the
suction vapor is unavoidable in an actual cycle,
whether or not a heat exchanger is used, any
practical means of causing the vapor to become
superheated in such a way that useful cooling
results are worthwhile. Hence, the value of a
heat exchanger lies in the fact that it provides a
method of superheating the vapor so that useful
cooling results. For this reason, the effect of a
heat exchanger on cycle efficiency can be
evaluated only by comparing the heat exchanger
cycle to one in which the vapor is superheated
without producing useful cooling.

The maximum amount of heat exchange
which can take place between the liquid and the
vapor in the heat exchanger depends on the
initial temperatures of the liquid and the vapor
as they enter the heat exchanger and on the
length of time they are in contact with each
other.

The greater the difference in temperature, the
greater is the exchange of heat for any given
period of contact. Thus, the lower the vaporiz-
ing temperature and the higher the condensing
temperature, the greater is the possible heat
exchange. Theoretically, if the two fluids
remained in contact for a sufficient length of
time, they would leave the heat exchanger at the
same temperature. In actual practice, this is
not possible. However, the longer the two
fluids stay in contact, the more nearly the two
temperatures will approach one another. Since
the specific heat of the vapor is less than that of
the liquid, the rise in the temperature of the

* It will be shown later that some advantages
accrue from superheating which takes place in the
compressor: (1) When the suction vapor absorbs
heat from the cylinder walls, the cylinder wall tem-
perature is lowered somewhat and this brings about
a desirable change in the path of the compression
process. However, the change is slight and is
difficult to evaluate. (2) When hermetic motor-
compressor assemblies are used, the suction vapor
should reach the compressor at a relatively low
temperature in order to help cool the motor windings.

ACTUAL REFRIGERATING CYCLES 117

vapor is always greater than the reduction in the
temperature of the liquid. For instance, the
specific heat of R-12 liquid is approximately
0.24 Btu per pound, whereas the specific heat of
the vapor is 0.15 Btu per pound. This means
that the temperature reduction of the liquid will
be approximately 62% (0.15/0.24) of the rise in
the temperature of the vapor, or that for each
24° F rise in the temperature of the vapor, the
temperature of the liquid will be reduced 15° F.

For the heat exchanger cycle in Fig. 8-9, the
vapor absorbs 5.71 Btu per pound in super-
heating from 20° F to 60° F. Assuming that all
of the superheating takes place in the heat
exchanger, the heat given up by the liquid is 5.71
Btu, so that the temperature of the liquid is
reduced 23.8° F (5.71/0.24) as the liquid passes
through the heat exchanger.
8-9. The Effect of Pressure Losses Resulting
from Friction. In overcoming friction, both
internal (within the fluid) and external (surface),
the refrigerant experiences a drop in pressure
while flowing through the piping, evaporator,
condenser, receiver, and through the valves and
passages of the compressor (Fig. 8-10).

A Ph diagram of an actual cycle, illustrating
the loss in pressure occurring in the various
parts of the system, is shown in Fig. 8-11. To
simplify the diagram, no superheating or sub-
cooling is shown and a simple saturated cycle is
drawn in for comparison.

Line J9'-C'represents the vaporizing process in
the evaporator during which the refrigerant
undergoes a drop in pressure of 5.5 psi. Whereas
the pressure and saturation temperature of the
liquid-vapor mixture at the evaporator inlet is
38.58 psia and 24° F, respectively, the pressure of
the saturated vapor leaving the evaporator is
33.08 psia, corresponding to a saturation tem-
perature of 16° F. The average vaporizing tem-
perature in the evaporator is 20° F, the same as
that of the saturated cycle.

As a result of the drop in pressure in the
evaporator, the vapor leaves the evaporator at a
lower pressure and saturation temperature and
with a greater specific volume than if no drop in
pressure occurred.

The refrigerating effect per pound and the
weight of refrigerant circulated per minute per
ton are approximately the same for both cycles,
but because of the greater specific volume the
volume of vapor handled by the compressor per

118 PRINCIPLES OF REFRIGERATION

38.58 psia

24* F (sat. temp.) *

Pressure drop through
evaporator, 5.5 psi

Average evaporating

temperature and

pressure
35.75 psia, 20° F

33.08 psia//
16* F (sat. temp.)

£L

, 158.9 psia

1 114"F (sat. temp.)
Pressure drop through

discharge valves, 8.2 psi

90° F (sat. temp.)

Pressure drop through
liquid line, 77.3 psi

150.7 psia

|TTlO°F (sat. temp.)

Pressure drop through

hot gas line and
condensers, 19.1 psi

— *^

\ 131.6 psia

100°F (sat. temp.)

Average condensing

temperature and pressure

139 psia, 104° F

Fig. 8-10. Flow diagram illustrating the effect of pressure drop in various parts of the system. Pressure drops
are exaggerated for clarity. (Refrigerant- 1 2.)

minute per ton is greater for the cycle experi-
encing the pressure drop. Too, because of the
lower pressure of the vapor leaving the evapora-
tor, the vapor must be compressed through a
greater pressure range during the compression
process, so that the horsepower per ton is also
greater for the cycle undergoing the drop in
pressure.

Line C'-C" represents the drop in pressure
experienced by the suction vapor in flowing

through the suction line from the evaporator to
the compressor inlet. Like pressure drop in the
evaporator, pressure drop in the suction line
causes the suction vapor to reach the compressor
at a lower pressure and in an expanded condition
so that the volume of vapor compressed per
minute per ton and the horsepower per ton are
both increased.

It is evident that the drop in pressure both in
the evaporator and in the suction line should be

Enthalpy (Btu/lb)

Pressure drop

1. Compressor discharge valves 4. Evaporator

2. Discharge line and condenser 5. Suction line

3. Liquid line 6. Compressor suction valves

Fig. 8-11. Ph diagram of refrig-
eration cycle illustrating the
effect of pressure losses in the
various parts of the system.
A simple saturated cycle is
drawn in for comparison.
(Refrigerant- 1 2.)

ACTUAL REFRIGERATING CYCLES 119

kept to an absolute minimum in order to obtain
the best possible cycle efficiency. This applies
also to heat exchangers or any other auxiliary
device intended for installation in the suction
line.

In Fig. 8-11, the pressure drops are exagger-
ated for clarity. Ordinarily, good evaporator
design limits the pressure drop across the
evaporator to 2 or 3 psi. Ideally, the suction line
should be designed so that the pressure drop is
between 1 and 2 psi.

against the spring-loading and to force the vapor
out through the discharge valves and passages
of the compressor into the discharge line.

Line D'-A represents the drop in pressure
resulting from the flow of the refrigerant through
the discharge line and condenser. That part of
line D'-A which represents the flow through the
discharge line will vary with the particular case,
since the discharge line may be either quite long
or very short, depending upon the application.
In any event, the result of the pressure drop will

Fig. 8-12. Ph diagram of actual
refrigeration cycle illustrating
effects of subcooling, super-
heating, and losses in pressure.
A simple saturated cycle is
drawn in for comparison.
(Refrigerant- 1 2).

Enthalpy (Btu/lb)

Pressure drop

1. Compressor discharge yalyes 4. Evaporator

2. Discharge line end condenser 5. Suction line

3. Liquid line 6. Compressor suction valves

Line C-C " represents the drop in pressure that
the suction vapor undergoes in flowing through
the suction valves and passages of the com-
pressor into the cylinder. The result of the drop
in pressure through the valves and passages on
the suction side of the compressor is the same as
if the drop occurred in the suction line, and the
effect on cycle efficiency is the same. Here again,
good design requires that the drop in pressure be
kept to a practical minimum.

Line C-D " represents the compression process
for the cycle undergoing the pressure drops.
Notice that the vapor in the cylinder is com-
pressed to a pressure considerably above the
average condensing pressure. It is shown later
that this is necessary in order to force the vapor
out of the cylinder through the discharge valves
against the condensing pressure and against the
additional pressure occasioned by the spring-
loading of the discharge valves.

Line D"-D' represents the drop in pressure
required to force the discharge valves open

be the same. Any drop in pressure occurring on
the discharge side of the compressor (in the dis-
charge valves and passages, in the discharge line,
and in the condenser) will have the effect of
raising the discharge pressure and thereby
increasing the work of compression and the
horsepower per ton.

Line A- A' represents the pressure drop result-
ing from the flow of the refrigerant through the
receiver tank and liquid line. Since the refrig-
erant at A' is a saturated liquid, the temperature
of the liquid must decrease as the pressure
decreases. If the liquid is not subcooled by
giving up heat to an external sink as its pressure
drops, a portion of the liquid must flash into a
vapor in the liquid line in order to provide the
required cooling of the liquid. Notice that point
A" lies in the region of phase-change, indicating
that a portion of the refrigerant is a vapor at
this point.

Despite the flashing of the liquid and the drop
in temperature coincident with the drop in

120 PRINCIPLES OF REFRIGERATION

pressure in the liquid line, the drop in pressure
in the liquid line has no effect on cycle efficiency.
The pressure and temperature of die liquid must
be reduced to the vaporizing condition before it
enters the evaporator in any case. The fact that
a part of this takes place in the liquid line rather
than in the refrigerant control has no direct
effect on the efficiency of the system. It does,
however, reduce the capacity of both the liquid
line and the refrigerant control. Furthermore,
passage of vapor through the refrigerant control
will eventually cause damage to the refrigerant
control by eroding the valve needle and seat.

Ordinarily, even without the use of a heat
exchanger, sufficient subcooling of the liquid
will occur in the liquid line to prevent the
flashing of the liquid if the drop in pressure in
the line is not excessive. Flashing of the liquid
in the liquid line will usually not take place when
the drop in the line does not exceed 5 psi.

The effect of pressure drop in the lines and in
the other parts of the system is discussed more
fully later in the appropriate chapters.

A Ph diagram of a typical refrigeratioircycle,
which illustrated the combined effects of pressure
drop, subcooling, and superheating, is compared
to the PA diagram of the simple saturated cycle
in Fig. 8-12.

PROBLEMS

1. The vaporizing and condensing temperature
of a Refrigerant-12 system are 40° F and 1 10° F,
respectively. The suction vapor is superheated

to 70° F in the suction line, whereas the liquid
is subcooled to 90° F by giving off heat to the
ambient air. Determine:

(a) The refrigerating effect per pound.

Ans. 54.01 Btu/lb

(b) The weight of refrigerant circulated per
minute per ton. Ans. 3.70 lb/min/ton

(d) The loss of refrigerating effect per pound
in the refrigerant control.

Ans. 11.7 Btu/lb

vapor. Ans. 4.39 Btu/lb

(/) The gain in refrigerating effect per pound

resulting from the liquid subcooling.

Ans. 4.93 Btu/lb
(g ) The adiabatic discharge temperature.

Ans. 138.5° F
(A) The heat of compression per pound.

Ans. 9 Btu/lb

(0 The heat of compression per minute per

ton. Ans. 33.3 Btu/min/ton

(J) The work of compression per minute per

ton. Ans. 25.907 lb/min/ton

(At) The theoretical horsepower per ton.

Ans. 0.755 hp/ton

(/)The heat rejected at the condenser per

pound. Arts. 67 .4 Btu/lb

(m) The heat rejected at the condenser per ton.

Ans. 249.38 Btu/min/ton

(») The coefficient of performance. Ans. 6

Note: Some of the properties of the refriger-
ant at various points in the cycle must be
determined from the Ph chart in Fig. 7-1.

Survey

of Refrigeration

Applications

9-1. History and Scope of the Industry. In

the early days of mechanical refrigeration, the
equipment available was bulky, expensive, and
not too efficient. Also it was of such a nature as
to require that a mechanic or operating engineer
be on duty at all times. This limited the use of
mechanical refrigeration to a few large applica-
tions such as ice plants, meat packing plants, and
large storage warehouses.

In the span of only a few decades refrigeration
has grown into the giant and rapidly expanding
industry that it is today. This explosive growth
came about as the result of several factors. First,
with the development of precision manufacturing
methods, it became possible to produce smaller,
more efficient equipment. This, along with the
development of "safe" refrigerants and the
invention of the fractional horsepower electric
motor, made possible the small refrigerating
unit which is so widely used at the present time
in such applications as domestic refrigerators
and freezers, small air conditioners, and com-
mercial fixtures. Today, there are few homes or
business establishments in the United States that
cannot boast of one or more mechanical
refrigeration units of some sort.

Few people outside of those directly connected
with the industry are aware of the significant
part that refrigeration has played in the develop-
ment of the highly technical society that

America is today, nor do they realize the extent
to which such a society is dependent upon
mechanical refrigeration for its very existence.
It would not be possible, for instance, to pre-
serve food in sufficient quantities to feed the
growing urban population without mechanical
refrigeration. Too, many of the large buildings
which house much of the nation's business and
industry would become untenable in the summer
months because of the heat if they were not air
conditioned with mechanical refrigerating
equipment.

In addition to the better known applications
of refrigeration, such as comfort air conditioning
and the processing, freezing, storage, transpor-
tation, and display of perishable products,
mechanical refrigeration is used in the processing
or manufacturing of almost every article or
commodity on the market today. The list of
processes or products made possible or improved
through the use of mechanical refrigeration is
almost endless. For example, refrigeration has
made possible the building of huge dams which
are vital to large-scale reclamation and hydro-
electric projects. It has made possible the con-
struction of roads and tunnels and the sinking of
foundation and mining shafts through and across
unstable ground formations. It has made
possible the production of plastics, synthetic
rubber, and many other new and useful mate-
rials and products. Because of mechanical
refrigeration, bakers can get more loaves of
bread from a barrel of flour, textile and paper
manufacturers can speed up their machines and
get more production, and better methods of
hardening steels for machine tools are available.
These represent only a few of the hundreds of
ways in which mechanical refrigeration is now
being used and many new uses are being found
each year. In fact, the only thing slowing the
growth of the refrigeration industry at the
present time is the lack of an adequate supply of
trained technical manpower.
9-2. Classification of Applications. For con-
venience of study, refrigeration applications may
be grouped into six general categories: (1)
domestic refrigeration, (2) commercial refrigera-
tion, (3) industrial refrigeration, (4) marine and
transportation refrigeration, (5) comfort air con-
ditioning, and (6) industrial air conditioning. It
will be apparent in the discussion which follows
that the exact limits of these areas are not

121

122 PRINCIPLES OF REFRIGERATION

precisely defined and that there is considerable
overlapping between the several areas.
9-3. Domestic Refrigeration. Domestic re-
frigeration is rather limited in scope, being
concerned primarily with household refrigera-
tors and home freezers. However, because the
number of units in service is quite large,
domestic refrigeration represents a significant
portion of the refrigeration industry.

Domesticunits are usually small in size, having
horsepower ratings of between ^ and J hp, and
are of the hermetically sealed type. Since these
applications are familiar to everyone, they will
not be described further here. However, the
problems encountered in the design and main-
tenance of these units are discussed in appro-
priate places in the chapters which follow.
9-4. Com.nercial Refrigeration. Commer-
cial refrigeration is concerned with the designing,
installation, and maintenance of refrigerated
fixtures of the type used by retail stores, res-
taurants, hotels, and institutions for the storing,
displaying, processing, and dispensing of perish-
able commodities of all types. Commercial
refrigeration fixtures are described in more
detail later in this chapter.
9-5. Industrial Refrigeration. Industrial
refrigeration is often confused with commercial
refrigeration because the division between these
two areas is not clearly defined. As a general
rule, industrial applications are larger in size
than commercial applications and have the dis-
tinguishing feature of requiring an attendant on
duty, usually a licensed operating engineer.
Typical industrial applications are ice plants,
large food-packing plants (meat, fish, poultry,
frozen foods, etc.), breweries, creameries, and
industrial plants, such as oil refineries, chemical
plants, rubber plants, etc. Industrial refrigera-
tion includes also those applications concerned
with the construction industry as described in
Section 9-1.

9-6. Marine and Transportation Refrigera-
tion. Applications falling into this category
could be listed partly under commercial refrig-
eration and partly under industrial refrigera-
tion. However, both these areas of specialization
have grown to sufficient size to warrant special
mention.

Marine refrigeration, of course, refers to
refrigeration aboard marine vessels and includes,
for example, refrigeration for fishing boats and

for vessels transporting perishable cargo as well
as refrigeration for the ship's stores on vessels of

all kinds.

Transportation refrigeration is concerned
with refrigeration equipment as it is applied to
trucks, both long distance transports and local
delivery, and to refrigerated railway cars. Typi-
cal refrigerated truck bodies are shown in Fig.

11-8.

9-7. Air Conditioning. As the name implies,
air conditioning is concerned with the condition
of the air in some designated area or space. This
usually involves control not only of the space
temperature but also of space humidity and air
motion, along with the filtering and cleaning of
the air.

Air conditioning applications are of two types,
either comfort or industrial, according to their
purpose. Any air conditioning which has as its
primary function the conditioning of air for
human comfort is called comfort air condition-
ing. Typical installations of comfort air
conditioning are in homes, schools, offices,
churches, hotels, retail stores, public buildings,
factories, automobiles, buses, trains, planes,
ships, etc.

On the other hand, any air conditioning which
does not have as its primary purpose the con-
ditioning of air for human comfort is called
industrial air conditioning. This does not
necessarily mean that industrial air conditioning
systems cannot serve as comfort air conditioning
coincidentally with their primary function.
Often this is the case, although not always so.
The applications of industrial air conditioning
are almost without limit both in number and in
variety. Generally speaking, the functions of
industrial air conditioning are to; (1) control
the moisture content of hydroscopic materials;
(2) govern the rate of chemical and biochemical
reactions; (3) limit the variations in the size of
precision manufactured articles because of ther-
mal expansion and contraction ; and (4) provide
clean, filtered air which is often essential to
trouble-free operation and to the production of
quality products.

9-8. Food Preservation. The preservation of
perishable commodities, particularly foodstuffs,
is one of the most common uses of mechanical
refrigeration. As such, it is a subject which
should be given consideration in any compre-
hensive study of refrigeration.

SURVEY OF REFRIGERATION APPLICATIONS 123

At the present time, food preservation is more
important than ever before in man's history.
Today's large urban populations require tremen-
dous quantities of food, which for the most part
must be produced and processed in outlying
areas. Naturally, these foodstuffs must be kept
in a preserved condition during transit and sub-
sequent storage until they are finally consumed.
This may be a matter of hours, days, weeks,
months, or even years in some cases. Too, many
products, particularly fruit and vegetables, are
seasonal. Since they are produced only during
certain seasons of the year, they must be stored
and preserved if they are to be made available
the year round.

As a matter of life or death, the preservation
of food has long been one of man's most pressing
problems. Almost from the very beginning of
man's existence on earth, it became necessary
for him to find ways of preserving food during
seasons of abundance in order to live through
seasons of scarcity. It is only natural, then, that
man should discover and develop such methods
of food preservation as drying, smoking, pick-
ling, and salting long before he had any know-
ledge of the causes of food spoilage. These
rather primitive methods are still widely used
today, not only in backward societies where no
other means are available but also in the most
modern societies where they serve to supplement
the more modern methods of food preservation.
For instance, millions of pounds of dehydrated
(dried) fruit, milk, eggs, fish, meat, potatoes, etc.,
are consumed in the United States each year,
along with huge quantities of smoked, pickled,
and salted products, such as ham, bacon, and
sausage, to name only a few. However, although
these older methods are entirely adequate for the
preservation of certain types of food, and often
produce very unusual and tasty products which
would not otherwise be available, they nonethe-
less have inherent disadvantages which limit
their usefulness. Since by their very nature they
bring about severe changes in appearance, taste,
and odor, which in many cases are objectionable,
they are not universally adaptable for the pres-
ervation of all types of food products. Further-
more, the keeping qualities of food preserved by
such methods are definitely limited as to time.
Therefore, where a product is to be preserved
indefinitely or for a long period of time, some
other means of preservation must be utilized.

The invention of the microscope and the sub-
sequent discovery of microorganisms as a major
cause of food spoilage led to the development of
canning in France during the time of Napoleon.
With the invention of canning, man found a way
to preserve food of all kinds in large quantities
and for indefinite periods of time. Canned foods
have the advantage of being entirely imperish-
able, easily processed, and convenient to handle
and store. Today, more food is preserved by
canning than by all other methods combined.
The one big disadvantage of canning is that
canned foods must be heat-sterilized, which
frequently results in overcooking. Hence, al-
though canned foods often have a distinctive and
delicious flavor all their own, they usually differ
greatly from the original fresh product.

The only means of preserving food in its
original fresh state is by refrigeration. This, of
course, is the principal advantage that refrigera-
tion has over other methods of food preser-
vation. However, refrigeration too has its
disadvantages. For instance, when food is to
be preserved by refrigeration, the refrigerating
process must begin very soon after harvesting
or killing and must be continuous until the food
is finally consumed. Since this requires rela-
tively expensive and bulky equipment, it is often
both inconvenient and uneconomical.

Obviously, then, there is no one method of
food preservation which is best in all cases and
the particular method used ih any one case will
depend upon a number of factors, such as the
type of product, the length of time the product
is to be preserved, the purpose for which the
product is to be used, the availability of trans-
portation and storage equipment, etc. Very
often it is necessary to employ several methods
simultaneously in order to obtain the desired
results.

9-9. Deterioration and Spoilage. Since the
preservation of food is simply a matter of pre-
venting or retarding deterioration and spoilage
regardless of the method used, a good knowledge
of the causes of deterioration and spoilage is
a prerequisite to the study of preservation
methods.

It should be recognized at the outset that
there are degrees of quality and that all perish-
able foods pass through various stages of
deterioration before becoming unfit for con-
sumption. In most cases, the objective in the

124 PRINCIPLES OF REFRIGERATION

preservation of food is not only to preserve the
foodstuff in an edible condition but also to
preserve it as nearly as possible at the peak of its
quality with respect to appearance, odor, taste,
and vitamin content. Except for a few processed
foods, this usually means maintaining the food-
stuff as nearly as possible in its original fresh
state.

Any deterioration sufficient to cause a detect-
able change in the appearance, odor, or taste of
fresh foods immediately reduces the commercial
value of the product and thereby represents an
economic loss. Consider, for example, wilted
vegetables or overripe fruit. Although their
edibility is little impaired, an undesirable change
in their appearance has been brought about
which usually requires that they be disposed of
at a reduced price. Too, since they are well on
their way to eventual spoilage, their keeping
qualities are greatly reduced and they must be
consumed or processed immediately or become
a total loss.

For obvious reasons, maintaining the vitamin
content at the highest possible level is always an
important factor in the processing and/or pres-
ervation of all food products. In fact, many
food processors, such as bakers and dairymen,
are now adding vitamins to their product to
replace those which are lost during processing.
Fresh vegetables, fruit, and fruit juices are some
of the food products which suffer heavy losses in
vitamin content very quickly if they are not
handled and protected properly. Although the
loss of vitamin content is not something which
in itself is apparent, in many fresh foods it is
usually accompanied by recognizable changes in
appearance, odor, or taste, such as, for instance,
wilting in leafy, green vegetables.

For the most part, the deterioration and even-
tual spoilage of perishable food are caused by a
series of complex chemical changes which take
place in the foodstuff after harvesting or killing.
These chemical changes are brought about by
both internal and external agents. The former
are the natural enzymes which are inherent in all
organic materials, whereas the latter are micro-
organisms which grow in and on the surface of
the foodstuff. Although either agent alone is
capable of bringing about the total destruction
of a food product, both agents are involved in
most cases of food spoilage. In any event, the
activity of both of these spoilage agents must be

either eliminated or effectively controlled if the
foodstuff is to be adequately preserved.
9-10. Enzymes. Enzymes are complex, pro-
tein-like, chemical substances. Not yet fully
understood, they are probably best described
as chemical catalytic agents which are capable
of bringing about chemical changes in organic
materials. There are many different kinds of
enzymes and each one is specialized in that it
produces only one specific chemical' reaction.
In general, enzymes are identified either by the
substance upon which they act or by the result
of their action. For instance, the enzyme,
lactase, is so known because it acts to convert
lactose (milk sugar) to lactic acid. This par-
ticular process is called lactic acid fermentation
and is the one principally responsible for the
"souring" of milk. Enzymes associated with
the various types of fermentation are sometimes
called ferments.

Essential in the chemistry of all living pro-
cesses, enzymes are normally present in all
organic materials (the cell tissue of all plants and
animals, both living and dead). They are manu-
factured by all living cells to help carry on the
various living activities of the cell, such as res-
piration, digestion, growth, and reproduction,
and they play an important part in such things
as the sprouting of seeds, the growth of plants
and animals, the ripening of fruit, and the
digestive processes of animals, including man.
However, enzymes are catabolic as well as ana-
bolic. That is, they act to destroy dead cell tissue
as well as to maintain live cell tissue. In fact,
enzymes are the agents primarily responsible
for the decay and decomposition of all organic
materials, as, for example, the putrification of
meat and fish and the rotting of fruit and
vegetables.

Whether their action is catabolic or anabolic,
enzymes are nearly always destructive to perish-
able foods. Therefore, except in those few
special cases where fermentation or putrification
is a part of the processing, enzymic action must
be either eliminated entirely or severely inhibited
if the product is to be preserved in good con-
dition. Fortunately, enzymes are sensitive to
the conditions of the surrounding media, par-
ticularly with regard to the temperature and the
degree of acidity or alkalinity, which provides a
practical means of controlling enzymic activity.
Enzymes are completely destroyed by high

SURVEY OF REFRIGERATION APPLICATIONS 125

temperatures that alter the composition of the
organic material in which they exist. Since most
enzymes are eliminated at temperatures above
160° F, cooking a food substance completely
destroys the enzymes contained therein. On the
other hand, enzymes are very resistant to low
temperatures and their activity may continue at
a slow rate even at temperatures below 0° F.
However, it is a well-known fact that the rate of
chemical reaction decreases as the temperature
decreases. Hence, although the enzymes are not
destroyed, their activity is greatly reduced at low
temperatures, particularly temperatures below
the freezing point of water.

Enzymic action is greatest in the presence of
free oxygen (as in the air) and decreases as the
oxygen supply diminishes.

With regard to the degree of acidity or alka-
linity, some enzymes require acid surroundings,
whereas others prefer neutral or alkaline en-
vironments. Those requiring acidity are de-
stroyed by alkalinity and those requiring
alkalinity are likewise destroyed by acidity.

Although an organic substance can be com-
pletely destroyed and decomposed solely by the
action of its own natural enzymes, a process
known as autolysis (self-destruction), this sel-
dom occurs. More often, the natural enzymes
are aided in their destructive action by enzymes
secreted by microorganisms.
9-11. Microorganisms. The term micro-
organism is used to cover a whole group of
minute plants and animals of microscopic and
submicroscopic size, of which only the following
three are of particular interest in the study of
food preservation: (1) bacteria, (2) yeasts, and
(3) molds. These tiny organisms are found in
large numbers everywhere — in the air, in the
ground, in water, in and on the bodies of plants
and animals, and in every other place where con-
ditions are such that living organisms can survive.

Because they secrete enzymes which attack
the organic materials upon which they grow,
microorganisms are agents of fermentation,
purification, and decay. As such, they are both
beneficial and harmful to mankind. Then-
growth in and on the surface of perishable foods
causes complex chemical changes in the food
substance which usually results in undesirable
alterations in the taste, odor, and appearance of
the food and which, if allowed to continue for
any length of time, will render the food unfit

for consumption. Too, some microorganisms
secrete poisonous substances (toxins) which are
extremely dangerous to health, causing poison-
ing, disease, and often death.

On the other hand, microorganisms have
many useful and necessary functions. As a
matter of fact, if it were not for the work of
microorganisms, life of any kind would not be
possible. Since decay and decomposition of all
dead animal tissue are essential to make space
available for new life and growth, the decaying
action of microorganisms is indispensable to the
life cycle.

Of all living things, only green plants (those
containing chlorophyll) are capable of using
inorganic materials as food for building their
cell tissue. Through a process called photo-
synthesis, green plants are able to utilize the
radiant energy of the sun to combine carbon
dioxide from the air with water and mineral
salts from the soil and thereby manufacture
from inorganic materials the organic compounds
which make up their cell tissue.

Conversely, all animals and all plants without
chlorophyll (fungi) require organic materials
(those containing carbon) for food to carry on
their life activities. Consequently, they must of
necessity feed upon the cell tissue of other plants
and animals (either living or dead) and are,
therefore, dependent either directly or indirectly
on green plants as a source of the organic
materials they need for life and growth.

It is evident, then, that should the supply of
inorganic materials in the soil, which serve as
food for green plants, ever become exhausted,
all life would soon disappear from the earth.
This is not likely to happen, however, since
microorganisms, as a part of their own living
process, are continuously replenishing the supply
of inorganic materials in the soil.

With the exception of a few types of soil
bacteria, all microorganisms need organic
materials as food to carry on the living process.
In most cases, they obtain these materials by
decomposing animal wastes and the tissue of
dead animals and plants. In the process of
decomposition, the complex organic compounds
which make up the tissue of animals and plants
are broken down step by step and are eventually
reduced to simple inorganic materials which are
returned to the soil to be used as food by the
green plants.

126 PRINCIPLES OF REFRIGERATION

In addition to the important part they play in
the "food chain" by helping to keep essential
materials in circulation, microorganisms ale
necessary in the processing of certain fermented
foods and other commodities. For example,
bacteria are responsible for the lactic acid fer-
mentation required in the processing of pickles,
olives, cocoa, coffee, sauerkraut, ensilage, and
certain sour milk products, such as butter,
cheese, buttermilk, yogurt, etc., and for the
acetic acid fermentation necessary in the pro-
duction of vinegar from various alcohols.
Bacterial action is useful also in the processing
of certain other commodities such as leather,
linen, hemp, and tobacco, and in the treatment
of industrial wastes of organic composition.

Yeasts, because of their ability to produce
alcoholic fermentation, are of immeasurable
value to the brewing and wine-making industries
and to the production of alcohols of all kinds.
Too, everyone is aware of the importance of
yeast in the baking industry.

The chief commercial uses of molds are in the
processing of certain types of cheeses and, more
important, in the production of antibiotics, such
as penicillin and aureomycin.

Despite their many useful and necessary
functions, the fact remains that microorganisms
are destructive to perishable foods. Hence,
their activity, like that of the natural enzymes,
must be effectively controlled if deterioration and
spoilage of the food substance are to be avoided.

Since each type of microorganism differs
somewhat in both nature and behavior, it is
worthwhile to examine each type separately.
9-12. Bacteria. Bacteria are a very simple
form of plant life, being made up of one single
living cell. Reproduction is accomplished by cell
division. On reaching maturity, the bacterium
divides into two separate and equal cells, each
of which in turn grows to maturity and divides
into two cells. Bacteria grow and reproduce at
an enormous rate. Under ideal conditions, a
bacterium can grow into maturity and repro-
duce in as little as 20 to 30 min. At this rate a
single bacterium is capable of producing as
many as 34,000,000,000,000 descendants in a
24-hr period. Fortunately, however, the life
cycle of bacteria is relatively short, being a
matter of minutes or hours, so that even under
ideal conditions they cannot multiply at any-
where near this rate.

The rate at which bacteria and other micro-
organisms grow and reproduce depends upon
such environmental conditions as temperature,
light, and the degree of acidity or alkalinity, and
upon the availability of oxygen, moisture, and
an adequate supply of soluble food. However,
there are many species of bacteria and they
differ greatly both in their choice of environment
and in the effect they have on their environment.
Like the higher forms of plant life, all species of
bacteria are not equally hardy with respect to
surviving adverse conditions of environment,
nor do all species thrive equally well under the
same environmental conditions. Some species
prefer conditions which are entirely fatal to
others. Too, some bacteria are spore-formers.
The spore is formed within the bacteria cell and
is protected by a heavy covering or wall. In the
spore state, which is actually a resting or
dormant phase of the organism, bacteria are
extremely resistant to unfavorable conditions of
environment and can survive in this state almost
indefinitely. The spore will usually germinate
whenever conditions become favorable for the
organism to carry on its living activities.

Most bacteria are saprophytes. That is, they
are "free living" and feed only on animal wastes
and on the dead tissue of animals and plants.
Some, however, are parasites and require a
living host. Most pathogenic bacteria (those
causing infection and disease) are of the para-
sitic type. In the absence of a living host, some
parasitic bacteria can live as saprophytes. Like-
wise, some saprophytes can live as parasites
when the need arises.

Since bacteria are not able to digest insoluble
food substances, they require food in a soluble
form. For this reason, most bacteria secrete
enzymes which are capable of rendering in-
soluble compounds into a soluble state, thereby
making these materials available to the bacteria
as food. The deterioration of perishable foods
by bacteria growth is a direct result of the action
of these bacterial enzymes.

Bacteria, like all other living things, require
moisture as well as food to carry on their life
activities. As in other things, bacteria vary
considerably in their ability to resist drought.
Although most species are readily destroyed by
drying and will succumb within a few hours, the
more hardy species are able to resist drought
for several days. Bacterial spores can withstand

SURVEY OF REFRIGERATION APPLICATIONS
The Growth of Bacteria in Milk in Various Periods

127

Time, hours

Temp., °F

168

2,400

2,100

1,850

1,400

2,500

3,600

218,000

4,200,000

3,100

12,000

1,480,000

11,600

540,000

180,000

28,000,000

1,400,000,000

Fig. 9-1. From ASRE Data Book, Applications Volume, 1956-57. Reproduced by permission of the American
Society of Heating, Refrigerating, and Air-Conditioning Engineers.

drought almost indefinitely, but will remain
dormant in the absence of moisture.

In their need for oxygen, bacteria fall into two
groups : (1) those which require free oxygen (air)
and (2) those which can exist without free oxy-
gen. Some species, although having a prefer-
ence for one condition or the other, can live
in the presence of free oxygen or in the absence
of it. Those bacteria living without free oxygen
obtain the needed oxygen through chemical
reaction which reduces one compound while
oxidizing another. Decomposition which occurs
in the presence of free oxygen is known as decay,
whereas decomposition which takes place in the
absence of free oxygen is called putrification.
One of the products of putrification is hydrogen
sulfide, a foul-smelling gas which is frequently
noted arising from decomposing animal
carcasses.

Bacteria are very sensitive to acidity or alka-
linity and cannot survive in an either highly acid
or highly alkaline environment. Most bacteria
require either neutral or slightly alkaline sur-
roundings, although some species prefer slightly
acid conditions. Because bacteria prefer neutral
or slightly alkaline surroundings, nonacid
vegetables are especially subject to bacterial
attack.

Light, particularly direct sunlight, is harmful
to all bacteria. Whereas visible light only
inhibits their growth, ultraviolet light is actually
fatal to bacteria. Since light rays, ultraviolet or
otherwise, have no power of penetration, they
are effective only in controlling surface bacteria.
However, ultraviolet radiation (usually from
direct sunlight), when combined with drying,
provides an excellent means of controlling
bacteria growth.

For each species of bacteria there is an opti-
mum temperature at which the bacteria will
grow at the highest rate. Too, for each species
there is a maximum and a minimum tempera-
ture which will permit growth. At temperatures
above the maximum, the bacteria are destroyed.
At temperatures below the minimum, the bac-
teria are rendered inactive or dormant. The
optimum temperature for most saprophytes is
usually between 75° F and 85° F, whereas the
optimum temperature for parasites is around
99° F or 100° F. A few species grow best at
temperatures near the boiling point of water,
whereas a few other types thrive best at tempera-
tures near the freezing point. However, most
species are either killed off or severely inhibited
at these temperatures. The effect of temperature
on the growth rate of bacteria is illustrated by
the chart in Fig. 9-1 which shows the growth
rate of bacteria in milk at various temperatures.
In general, the growth rate of bacteria is con-
siderably reduced by lowering the temperature.
9-13. Yeasts. Yeasts are simple, one-cell plants
of the fungus family. Of microscopic size,
yeast cells are somewhat larger and more com-
plex than the bacteria cells. Although a few
yeasts reproduce by fission or by sexual process,
reproduction is usually by budding. Starting as
a small protrusion of the mature cell, the bud
enlarges and finally separates from the mother
cell. Under ideal conditions, budding is fre-
quently so rapid that new buds are formed before
separation occurs so that yeast clusters are
formed.

Like bacteria, yeasts are agents of fermenta-
tion and decay. They secrete enzymes that bring
about chemical changes in the food upon which
they grow. Yeasts are noted for their ability

128 PRINCIPLES OF REFRIGERATION

to transform sugars into alcohol and carbon
dioxide. Although destructive to fresh foods,
particularly fruits and berries and their juices,
the alcoholic fermentation produced by yeasts
is essential to the baking, brewing, and wine-
making industries.

Yeasts are spore-formers, with as many as
eight spores being formed within a single yeast
cell. Yeasts are widespread in nature and yeast
spores are invariably found in the air and on the
skin of fruit and berries, for which they have a
particular affinity. They usually spend the
winter in the soil and are carried to the new fruit
in the spring by insects or by the wind.

Like bacteria, yeasts require air, food, and
moisture for growth, and are sensitive to tem-
perature and the degree of acidity or alkalinity
in the environment. For the most part, yeasts
prefer moderate temperatures and slight acidity.
In general, yeasts are not as resistant to unfavor-
able conditions as are bacteria, although they
can grow in acid surroundings which inhibit
most bacteria. Yeast spores, like those of bac-
teria, are extremely hardy and can survive for
long periods under adverse conditions.
9-14. Molds. Molds, like yeasts, are simple
plants of the fungi family. However, they are
much more complex in structure than either
bacteria or yeasts. Whereas the individual
bacteria or yeast plants consist of one single
cell, an individual mold plant is made up of a
number of cells which are positioned end to end
to form long, branching, threadlike fibers called
hypha. The network which is formed by a mass
of these threadlike fibers is called the mycelium
and is easily visible to the naked eye. The
hyphae of the mold plant are of two general
types. Some are vegetative fibers which grow
under the surface and act as roots to gather food
for the plant, whereas others, called aerial
hyphae, grow on the surface and produce the
fruiting bodies.

Molds reproduce by spore formation. The
spores develop in three different ways, depending
on the type of mold: (1) as round clusters
within the fibrous hyphae network, (2) as a mass
enclosed in a sac and attached to the end of
aerial hyphae, and (3) as chainlike clusters on
the end of aerial hyphae. In any case, a single
mold plant is capable of producing thousands of
spores which break free from the mother plant
and float away with the slightest air motion.

Mold spores are actually seeds and, under the
proper conditions, will germinate and produce
mold growth on any food substance with which
they come in contact. Since they are carried
about by air currents, mold spores are found
almost everywhere and are particularly abundant
in the air.

Although molds are less resistant to high
temperatures than are bacteria, they are more
tolerant to low temperatures, growing freely at
temperatures close to the freezing point of water.
Mold growth is inhibited by temperatures below
32° F, more from the lack of free moisture than
from the effect of low temperature. All mold
growth ceases at temperatures of 10° F and
below.

Molds flourish in dark, damp surroundings,
particularly in still air. An abundant supply of
oxygen is essential to mold growth, although a
very few species can grow in the absence of
oxygen. Conditions inside cold-storage rooms
are often ideal for mold growth, especially in
the wintertime. This problem can be overcome
somewhat by maintaining good air circulation
in the storage room, by the use of germicidal
paints, and ultraviolet radiation, and by frequent
scrubbing.

Unlike bacteria, molds can thrive on foods
containing relatively large amounts of sugars or
acids. They are often found growing on acid
fruits and on the surface of pickling vats, and
are the most common cause of spoilage in
apples and citrus fruits.

9-15. Control of Spoilage Agents. Despite
complications arising from the differences in the
reaction of the various types of spoilage agents
to specific conditions in the environment, con-
trolling these conditions provides the only means
of controlling these spoilage agents. Thus, all
methods of food preservation must of necessity
involve manipulation of the environment in and
around the preserved product in order to pro-
duce one or more conditions unfavorable to the
continued activity of the spoilage agents. When
the product is to be preserved for any length of
time, the unfavorable conditions produced must
be of sufficient severity to eliminate the spoilage
agents entirely or at least render them ineffective
or dormant.

All types of spoilage agents are destroyed when
subjected to high temperatures over a period of
time. This principle is used in the preservation

SURVEY OF REFRIGERATION APPLICATIONS 129

of food by canning. The temperature of the
product is raised to a level fatal to all spoilage
agents and is maintained at this level until they
are all destroyed. The product is then sealed
in sterilized, air-tight containers to prevent
recontamination. A product so processed will
remain in a preserved state indefinitely.

The exposure time required for the destruc-
tion of all spoilage agents depends upon the
temperature level. The higher the temperature
level, the shorter is the exposure period required.
In this regard, moist heat is more effective than
dry heat because of its greater penetrating
powers. When moist heat is used, the tempera-
ture level required is lower and the processing
period is shorter. Enzymes and all living micro-
organisms are destroyed when exposed to the
temperature of boiling water for approximately
five minutes, but the more resistant bacteria
spores may survive at this condition for several
hours before succumbing. For this reason, some
food products, particularly meats and nonacid
vegetables, require long processing periods
which frequently result in overcooking of the
product. These products are usually processed
under pressure so that the processing tempera-
ture is increased and the processing time
shortened.

Another method of curtailing the activity of
spoilage agents is to deprive them of the mois-
ture and/or food which is necessary for then-
continued activity. Both enzymes and micro-
organisms require moisture to carry on their
activities. Hence, removal of the free moisture
from a product will severely limit their activities.
The process of moisture removal is called drying
(dehydration) and is one of the oldest methods of
preserving foods. Drying is accomplished either
naturally in the sun and air or artificially in
ovens. Dried products which are stored in a
cool, dry place will remain in good condition
for long periods.

Pickling is essentially a fermentation process,
the end result of which is the exhaustion of the
substances which serve as food for yeasts and
bacteria. The product to be preserved by pick-
ling is immersed in a salt brine solution and
fermentation is allowed to take place, during
which the sugars contained in the food product
are converted to lactic acid, primarily through
the action of lactic acid bacteria.
Smoked products are preserved partially by

the drying effect of the smoke and partially
by antiseptics (primarily creosote) which are
absorbed from the smoke.

Too, some products are "cured" with sugar
or salt which act as preservatives in that they
create conditions unfavorable to the activity of
spoilage agents. Some other frequently used
preservatives are vinegar, borax, saltpeter, bon-
zoate of soda, and various spices. A few of the
products preserved in this manner are sugar-
cured hams, salt pork, spiced fruits, certain
beverages, jellies, jams, and preserves.
9-16. Preservation by Refrigeration. The
preservation of perishables by refrigeration in-
volves the use of low temperature as a means of
eliminating or retarding the activity of spoilage
agents. Although low temperatures are not as
effective in bringing about the destruction of
spoilage agents as are high temperatures, the
storage of perishables at low temperatures
greatly reduces the activity of both enzymes and
microorganisms and thereby provides a prac-
tical means of preserving perishables in their
original fresh state for varying periods of time.
The degree of low temperature required for
adequate preservation varies with the type of
product stored and with the length of time the
product is to be kept in storage.

For purposes of preservation, food products
can be grouped into two general categories : (1)
those which are alive at the time of distribution
and storage and (2) those which are not. Non-
living food substances, such as meat, poultry,
and fish, are much more susceptible to micro-
bial contamination and spoilage than are living
food substances, and they usually require more
stringent preservation methods.

With nonliving food substances, the problem
of preservation is one of protecting dead tissue
from all the forces of putrification and decay,
both enzymic and microbial. In the case of
living food substances, such as fruit and vege-
tables, the fact of life itself affords considerable
protection against microbial invasion, and the
preservation problem is chiefly one of keeping
the food substance alive while at the same time
retarding natural enzymic activity in order to
slow the rate of maturation or ripening.

Vegetables and fruit are as much alive after
harvesting as they are during the growing period.
Previous to harvesting they receive a continuous
supply of food substances from the growing

130 PRINCIPLES OF REFRIGERATION

plant, some of which is stored in the vegetable
or fruit. After harvesting, when the vegetable or
fruit is cut off from its normal supply of food,
the living processes continue through utilization
of the previously stored food substances. This
causes the vegetable or fruit to undergo
changes which will eventually result in deteriora-
tion and complete decay of the product. The
primary purpose of placing such products under
refrigeration is to slow the living processes by
retarding enzymic activity, thereby keeping the
product in a preserved condition for a longer
period.

Animal products (nonliving food substances)
are also affected by the activity of natural
enzymes. The enzymes causing the most
trouble are those which catalyze hydrolysis and
oxidation and are associated with the break-
down of animal fats. The principal factor
limiting the storage life of animal products, in
both the frozen and unfrozen states, is rancidity.
Rancidity is caused by oxidation of animal fats
and, since some types of animal fats are less
stable than others, the storage life of animal
products depends in part on fat composition.
For example, because of the relative stability of
beef fat, the storage life of beef is considerably
greater than that of pork or fish whose fatty
tissues are much less stable.

Oxidation and hydrolysis are controlled by
placing the product under refrigeration so that
the activity of the natural enzymes is reduced.
The rate of oxidation can be further reduced in
the case of animal products by packaging the
products in tight-fitting, gas-proof containers
which prevent air (oxygen) from reaching the
surface of the product. The packaging of fruit
and vegetables in gas-proof containers, when
stored in the unfrozen state, is not practical.
Since these products are alive, packaging in gas-
proof containers will cause suffocation and
death. A dead fruit or vegetable decays very
quickly.

As a general rule, the lower the storage tem-
perature, the longer is the storage life of the
product.

9-17. Refrigerated Storage. Refrigerated
storage may be divided into three general cate-
gories: (1) short-term or temporary storage,
(2) long-term storage, and (3) frozen storage.
For short- and long-term storage, the product is
chilled and stored at some temperature above

its freezing point, whereas frozen storage re-
quires freezing of the product and storage at
some temperature between 10° F and — 10° F,
with 0° F being the temperature most frequently
employed.

Short-term or temporary storage is usually
associated with retail establishments where
rapid turnover of the product is normally ex-
pected. Depending upon the product, short-
term storage periods range from one or two days
in some cases to a week or more in others, but
seldom for more than fifteen days.

Long-term storage is usually carried out by
wholesalers and commercial storage warehouses.
Again, the storage period depends on the type
of product stored and upon the condition of the
product on entering storage. Maximum storage
periods for long-term storage range from seven
to ten days for some sensitive products, such as
ripe tomatoes, cantaloupes, and broccoli, and up
to six or eight months for the more durable
products, such as onions and some smoked
meats. When perishable foods are to be stored
for longer periods, they should be frozen and
placed in frozen storage. Some fresh foods,
however, such as tomatoes, are damaged by the
freezing process and therefore cannot be success-
fully frozen. When such products are to be
preserved for long periods, some other method
of preservation should be used.
9-18. Storage Conditions. The optimum
storage conditions for a product held in either
short- or long-term storage depends upon the
nature of the individual product, the length of
time the product is to be held in storage, and
whether the product is packaged or unpackaged.
In general, the conditions required for short-
term storage are more flexible than those
required for long-term storage and, ordinarily,
higher storage temperatures are permissible.
Recommended storage conditions for both
short- and long-term storage and the approxi-
mate storage life for various products are listed
in Tables 10-10 through 10-13, along with other
product data. These data are the result of both
experiment and experience and should be
followed closely, particularly for long-term
storage, if product quality is to be maintained
at a high level during the storage period.
9-19. Storage Temperature. Examination of
the tables will show that the optimum storage
temperature for most products is one slightly

SURVEY OF REFRIGERATION APPLICATIONS 131

above the freezing point of the product. There
are, however, notable exceptions.

Although the effect of incorrect storage tem-
peratures generally is to lower product quality
and shorten storage life, some fruits and vege-
tables are particularly sensitive to storage tem-
perature and are susceptible to so-called cold
storage diseases when stored at temperatures
above or below their critical storage tempera-
tures. For example, citrus fruits frequently
develop rind pitting when stored at relatively
high temperatures. On the other hand, they are
subject to scald (browning of the rind) and
watery breakdown when stored at temperatures
below their critical temperature. Bananas suffer
peel injury when stored below 56° F, whereas
celery undergoes soggy breakdown when stored
at temperatures above 34° F. Although onions
tend to sprout at temperatures above 32° F,
Irish potatoes tend to become sweet at storage
temperatures below 40° F. Squash, green beans,
and peppers develop pits on their surface when
stored at or near 32° F. Too, whereas the best
storage temperature for most varieties of apples
is 30° F to 32° F, some varieties are subject to
soft scald and soggy breakdown when stored
below 35° F. Others develop brown core at
temperatures below 36° F, and still others
develop internal browning when stored below
40° F.

9-20. Humidity and Air Motion. The
storage of all perishables in their natural state
(unpackaged) requires close control not only of
the space temperature but also of space humid-
ity and air motion. One of the chief causes of
the deterioration of unpackaged fresh foods,
such as meat, poultry, fish, fruit, vegetables,
cheese, eggs, etc., is the loss of moisture from
the surface of the product by evaporation into
the surrounding air. This process is known as
desiccation or dehydration. In fruit and vege-
tables, desiccation is accompanied by shriveling
and wilting and the product undergoes a
considerable loss in both weight and vitamin
content. In meats, cheese, etc., desiccation
causes discoloration, shrinkage) and heavy
trim losses. It also increases the rate of oxida-
tion. Eggs lose moisture through the porous
shell, with a resulting loss of weight and general
downgrading of the egg.

Desiccation will occur whenever the vapor
pressure of the product is greater than the vapor

pressure of the surrounding air, the rate of
moisture loss from the product being propor-
tional to the difference in the vapor pres-
sures and to the amount of exposed product
surface.

The difference in vapor pressure between the
product and the air is primarily a function of
the relative humidity and the velocity of the air
in the storage space. In general, the lower the
relative humidity and the higher the air velocity,
the greater will be the vapor pressure differential
and the greater the rate of moisture loss from
the product. Conversely, minimum moisture
losses are experienced when the humidity in the
storage space is maintained at a high level with
low air velocity. Hence, 100% relative humidity
and stagnant air are ideal conditions for pre-
venting dehydration of the stored product.
Unfortunately, these conditions are also con-
ducive to rapid mold growth and the formation
of slime (bacterial) on meats. Too, good
circulation of the air in the refrigerated space
and around the product is necessary for ade-
quate refrigeration of the product. For these
reasons, space humidity must be maintained at
somewhat less than 100% and air velocities
must be sufficient to provide adequate air
circulation. The relative humidities and air
velocities recommended for the storage of
various products are listed in Tables 10-10
through 10-13.

When the product is- stored in vapor-proof
containers, space humidity and air velocity are
not critical. Some products, such as dried
fruits, tend to be hydroscopic and therefore
require storage at low relative humidities.
9-21. Mixed Storage. Although the main-
tenance of optimum storage conditions requires
separate storage facilities for most products,
this is not usually economically feasible. There-
fore, except when large quantities of product
are involved, practical considerations often
demand that a number of refrigerated products
be placed in common storage. Naturally, the
difference in the storage conditions required by
the various products raises a problem with
regard to the conditions to be maintained in a
space designed for common storage.

As a general rule, storage conditions in such
spaces represent a compromise and usually
prescribe a storage temperature somewhat above
the optimum for some of the products held in

132 PRINCIPLES OF REFRIGERATION

mixed storage. The higher storage tempera-
tures are used in mixed storage in order to
minimize the chances of damaging the more
sensitive products which are subject to the
aforementioned "cold storage diseases" when
stored at temperatures below their critical
temperature.

Although higher storage temperatures tend
to shorten the storage life of some of the products
held in mixed storage, this is not ordinarily a
serious problem when the products are stored
only for short periods as in temporary
storage.

For long-term storage, most of the larger
wholesale and commercial storage warehouses
have a number of separate storage spaces
available. General practice in such cases is to
group the various products for storage, and
only those products requiring approximately
the same storage conditions are placed together
in common storage.

Another problem associated with mixed
storage is that of odor and flavor absorption.
Some products absorb and/or give off odors
while in storage. Care should be taken not to
store such products together even for short
periods. Dairy products in particular are very
sensitive with regard to absorbing odors and
flavors from other products held in mixed
storage. On the other hand, potatoes are
probably the worst offenders in imparting off-
flavors to other products in storage and should
never be stored with fruit, eggs, dairy products,
or nuts.

9-22. Product Condition on Entering Stor-
age. One of the principal factors determining
the storage life of a refrigerated product is the
condition of the product on entering storage.
It must be recognized that refrigeration merely
arrests or retards the natural processes of
deterioration and therefore cannot restore to
good condition a product which has already
deteriorated. Neither can it make a high
quality product out of one of initial poor
quality. Hence, only vegetables and fruit in
good condition should be accepted for storage.
Those that have been bruised or otherwise
damaged, particularly if the skin has been
broken, have lost much of their natural protec-
tion against microbial invasion and are there-
fore subject to rapid spoilage by these agents.
Too, as a general rule, since maturation and

ripening continue after harvesting, vegetables
and fruit intended for storage should be
harvested before they are fully mature. The
storage life of fully mature or damaged fruit
and vegetables is extremely short even under
the best storage conditions, and such products
should be sent directly to market to avoid
excessive losses.

Since" a food product begins to deteriorate
very quickly after harvesting or killing, it is
imperative that preservation measures be taken
immediately. To assure maximum storage life
with minimum loss of quality, the product should
be chilled to the storage temperature as soon
as possible after harvesting or killing. When
products are to be shipped over long distances
to storage, they should be precooled and shipped
by refrigerated transport.
9-23. Product Chilling. Product chilling is
distinguished from product storage in that the
product enters the chilling room at a high
temperature (usually either harvesting or killing
temperature) and is chilled as rapidly as possible
to the storage temperature, whereupon it is
normally removed from the chilling room and
placed in a holding cooler for storage. The
handling of the product during the chilling
period has a marked influence on the ultimate
quality and storage life of the product.

The recommended conditions for product
chilling rooms are given in Tables 10-10 through
10-13. Before the hot product is loaded into
the chilling room, the chilling room temperature
should be at the "chill finish" temperature.
During loading and during the early part of the
chilling period, the temperature and vapor
pressure differential between the product and
the chill room air will be quite large and the
product will give off heat and moisture at a
high rate. At this time, the temperature and
humidity in the chill room will rise to a peak as
indicated by the "chill start" conditions in the
tables.* At the end of the cycle, the chill room
temperature will again drop to the "chill finish"
conditions. It is very important that the
refrigerating equipment have sufficient capacity

* The temperatures listed in the tables as chill
start temperatures are average values and are in-
tended for use in selecting the refrigerating equip-
ment. Actual temperatures in the chilling room
during the peak chilling period are usually 3° F to
4° F higher than those listed.

SURVEY OF REFRIGERATION APPLICATIONS 133

to prevent the chill room temperature from
rising excessively during the peak chilling period.
9-24. Relative Humidity and Air Velocity
in Chill Rooms. The importance of relative
humidity in chilling rooms depends upon the
product being chilled, particularly upon whether
the product is packaged or not. Naturally,
when the product is chilled in vapor-proof
containers, the humidity in the chilling room
is relatively unimportant. However, during
loading and during the initial stages of chilling,
chilling room humidity will be high if the con-
tainers are wet, but will drop rapidly once the
free moisture has been evaporated.

Products chilled in their natural state (un-
packaged) lose moisture very rapidly, often
producing fog in the chilling room during the
early stages of chilling when the product tem-
perature and vapor pressure are high. Rapid
chilling and high air velocity are desirable
during this time so that the temperature and
vapor pressure of the product are lowered as
quickly as possible in order to avoid excessive
moisture loss and shrinkage. High air velocity
is needed also in order to carry away the vapor
and thereby prevent condensation of moisture
on the surface of the product.

Although high air velocity tends to increase
the rate of evaporation of moisture from the
product, it also greatly accelerates the chilling
rate and results in a more rapid reduction in
product temperature and vapor pressure. Since
the reduction in vapor pressure resulting from
the higher chilling rate more than offsets the
increase in the rate of evaporation occasioned
by the higher air velocity, the net effect of the
higher air velocity during the early stages of
chilling is to reduce the over-all loss of moisture
from the product. However, during the final
stages of chilling, when the temperature and
vapor pressure of the product are considerably
lower, the effect of high air velocity in the chill-
ing room is to increase the rate of moisture loss
from the product. Therefore, the air velocity
in the chilling room should be reduced during
the final stages of chilling.

As a general rule, the humidity should be
kept at a high level when products subject to
dehydration are being chilled. Some extremely
sensitive products, such as poultry and fish,
are frequently chilled in ice slush to reduce
moisture losses during chilling. For the same

reason, eggs are sometimes dipped in a light
mineral oil before chilling and storage. Too,
poultry, fish, and some vegetables are often
packed in ice for chilling and storage. When
products packed in ice are placed in refrigerated
storage, the slowly melting ice keeps the surface
of the product moist and prevents excessive
dehydration.

9-25. Freezing and Frozen Storage. When a
product is to be preserved in its original fresh
state for relatively long periods, it is usually
frozen and stored at approximately 0° F or
below. The list of food products commonly
frozen includes not only those which are
preserved in their fresh state, such as vegetables,
fruit, fruit juices, berries, meat, poultry, sea
foods, and eggs (not in shell), but also many
prepared foods, such as breads, pastries, ice
cream, and a wide variety of specially prepared
and precooked food products, including full
dinners.

The factors governing the ultimate quality
and storage life of any frozen product are:

1 . The nature and composition of the product
to be frozen

2. The care used in selecting, handling, and
preparing the product for freezing

3. The freezing method

4. The storage conditions.

Only high quality products in good condition
should be frozen. With vegetables and fruit,
selecting the proper variety for freezing is very
important. Some varieties are not suitable for
freezing and will result in a low quality product
or in one with limited keeping qualities.

Vegetables and fruit to be frozen should be
harvested at the peak of maturity and should be
processed and frozen as quickly as possible
after harvesting to avoid undesirable chemical
changes through enzymic and microbial action.

Both vegetables and fruit require considerable
processing before freezing. After cleaning and
washing to remove foreign materials — leaves,
dirt, insects, juices, etc. — from their surfaces,
vegetables are "blanched" in hot water or
steam at 212° F in order to destroy the natural
enzymes. It will be remembered that enzymes
are not destroyed by low temperature and,
although greatly reduced, their activity con-
tinues at a slow rate even in food stored at 0° F
and below. Hence, blanching, which destroys

134 PRINCIPLES OF REFRIGERATION

Fig. 9-2. Walk-in installation. Suspended blast
freezer provides high-velocity air for fast freezing,
saving valuable floor space in small areas. (Courtesy
Carrier Corporation.)

Fig. 9-3. Suspended blast free-
zer applied to reach-in cabinet
distributes blast air through
shelves. (Courtesy Carrier
Corporation.)

Fig. 9-4. Freezing in one room
and storage in another is
accomplished by single, floor-
mounted blast freezers. (Cour-
tesy Carrier Corporation.)

SURVEY OF REFRIGERATION APPLICATIONS 135

most of the enzymes, greatly increases the
storage life of frozen vegetables. The time
required for blanching varies with the type and
variety of the vegetable and ranges from 1 to l£
min for green beans to 1 1 min for large ears of
corn. Although much of the microbial popula-
tion is destroyed along with the enzymes during
the blanching process, many bacteria survive.
To prevent spoilage by these viable bacteria,
vegetables should be chilled to 50° F immedi-
ately after blanching and before they are pack-
aged for the freezer.

Like vegetables, fruit must also be cleaned
and washed to remove foreign materials and
to reduce microbial contamination. Although
fruit is perhaps even more subject to enzymic
deterioration than are vegetables, it is never
blanched to destroy the natural enzymes since
to do so would destroy the natural fresh quality
which is so desirable.

The enzymes causing the most concern with
regard to frozen fruit are the ones which
catalyze oxidation and result in rapid browning
of the flesh. To control oxidation, fruit to be
frozen is covered with a light sugar syrup. In
some cases, ascorbic acid, citric acid, or sulfur
dioxide are also used for this purpose.

As a general rule, meat products do not
require any special processing prior to freezing.
However, because of consumer demand, speci-
ally prepared meats and meat products are being
frozen in increasing amounts. This is true also
of poultry and sea foods.

Because of the relative instability of their
fatty tissue, pork and fish are usually frozen as
soon after chilling as possible. On the other
hand, beef is frequently "aged" in a chilling
cooler for several days before freezing. During
this time the beef is tenderized to some extent
by enzymic activity. However, the aging of
beef decreases its storage life, particularly if the
aging period exceeds 6 or 7 days.

With poultry, experiments indicate that
poultry frozen within 12 to 24 hr after killing
is more tender than that frozen immediately
after killing. However, delaying freezing beyond
24 hr tends to reduce storage life without
appreciable increasing tenderness.
9-26. Freezing Methods. Food products may
be either sharp (slow) frozen or quick frozen.
Sharp freezing is accomplished by placing the
product in a low temperature room and allowing

it to freeze slowly, usually in still air. The tem-
perature maintained in sharp freezers ranges
from 0° F to —40° F. Since air circulation is
usually by natural convection, heat transfer
from the product ranges from 3 hr to 3 days,
depending upon the bulk of the product and
upon the conditions in the sharp freezer. Typi-
cal items which are sharp frozen are beef and
pork half-carcasses, boxed poultry, panned and
whole fish, fruit in barrels and other large
containers, and eggs (whites, yolks, or whole)
in 10 and 30 lb cans.

Quick freezing is accomplished in any one or
in any combination of three ways: (1) immer-
sion, (2) indirect contact, and (3) air blast.
9-27. Air Blast Freezing. Air blast freezing
utilizes the combined effects of low temperature
and high air velocity to produce a high rate of
heat transfer from the product. Although the
method employed varies considerably with the
application, blast freezing is accomplished by
circulating high-velocity, low-temperature air
around the product. Regardless of the method
used, it is important that the arrangement of
the freezer is such that air can circulate freely
around all parts of the product.

Packaged blast freezers are available in both
suspended and floor-mounted models. Typical
applications are shown in Figs. 9-2 through 9-4.
Blast freezing is frequently carried out in
insulated tunnels, particularly where large
quantities of product are to be frozen (Figs.
9-5 and 9-6). In some instances, the product
is carried through the freezing tunnel and
frozen on slow-moving, mesh conveyor belts.
The unfrozen product is placed on the con-
veyor at one end of the tunnel and is frozen
by the time it reaches the other end. Another
method is to load the product on tiered dollies.
The dollies are pushed into the tunnel and the
product is frozen; whereupon they are pushed
out of the freezing tunnel into a storage room
(Fig. 9-5).

Although blast freezing is used to freeze
nearly all types of products, it is particularly
suitable for freezing products of nonuniform
or irregular sizes and shapes, such as dressed
poultry.

9-28. Indirect Contact Freezing. Indirect
freezing is usually accomplished in plate freezers
and involves placing the product on metal
plates through which a refrigerant is circulated

136 PRINCIPLES OF REFRIGERATION

Fig. 9-5. Packaged blast freezers
applied to freezing tunnel. High
velocity, — 15° F air is blasted
through trucks. (Courtesy Car-
rier Corporation.)

(Fig. 9-7). Since the product is in direct thermal
contact with the refrigerated plate, heat transfer
from the product occurs primarily by conduction
so that the efficiency of the freezer depends, for
the most part, on the amount of contact surface.
This type of freezer is particularly useful when
products are frozen in small quantities.

One type of plate freezer widely used by the
larger commercial freezers to handle small,
flat, rectangular, consumer-size packages is the
multiplate freezer. The multiplate freezer
consists of a series of horizontal, parallel,
refrigerated plates which are actuated by
hydraulic pressure so that they can be opened
to receive the product between them and then
closed on the product with any desired pressure.
When the plates are closed, the packages are
held tightly between the plates. Since both the
top and the bottom of the packages are in good
thermal contact with the refrigerated plates,
the rate of heat transfer is high and the product
is quickly frozen.

9-29. Immersion Freezing. Immersion freez-
ing is accomplished by immersing the product in
a low temperature brine solution, usually either
sodium chloride or sugar. Since the refrigerated
liquid is a good conductor and is in good
thermal contact with all the product, heat
transfer is rapid and the product is completely
frozen in a very short time.

Another advantage of immersion freezing is
that the product is frozen in individual units
rather than fused together in a mass.

The principal disadvantage of immersion

freezing is that juices tend to be extracted from
the product by osmosis. This results in con-
tamination and weakening of the freezing
solution. Too, where a sodium chloride brine
is used, salt penetration into the product may
sometimes be excessive. On the other hand,
when fruit is frozen in a sugar solution, sugar
penetration into the fruit is entirely beneficial.

The products most frequently frozen by
immersion are fish and shrimp. Immersion is
particularly suitable for freezing fish and
shrimp at sea, since the immersion freezer is
relatively compact and space aboard ship is at
a premium. In addition, immersion freezing
produces a "glaze" (thin coating of ice) on the
surface of the product which helps to prevent
dehydration of unpackaged products during
the storage period.

9-30. Quick Freezing vs. Sharp Freezing.
Quick frozen products are nearly always
superior to those which are sharp (slow) frozen.
D. K. Tressler, in 1932, summarized the views
of R. Plank, H. F. Taylor, C. Birdseye, and
G. A. Fitzgerald, and stated the following as the
main advantages of quick freezing over slow
freezing:

1 . The ice crystals formed are much smaller,
and therefore cause much less damage to cells.

2. The freezing period being much shorter,
less time is allowed for the diffusion of salts and
the separation of water in the form of ice.

3. The product is quickly cooled below the
temperature at which bacterial, mold, and yeast

SURVEY OF REFRIGERATION APPLICATIONS 137

o
u

138 PRINCIPLES OF REFRIGERATION

Fig. 9-7. Plate freezer for indirect
contact freezing. (Courtesy Dole
Refrigerating Company.)

growth occurs, thus preventing decomposition
during freezing.*

The principal difference between quick freez-
ing and sharp freezing is in the size, number,
and location of the ice crystals formed in the
product as cellular fluids are solidified. When
a product is slow frozen, large ice crystals are
formed which result in serious damage to the
tissue of some products through cellular break-
down. Quick freezing, on the other hand,
produces smaller ice crystals which are formed
almost entirely within the cell so that cellular
breakdown is greatly reduced. Upon thawing,
products which have experienced considerable
cellular damage are prone to lose excessive
amounts of fluids through "drip" or "bleed,"
with a resulting loss of quality.

Ice-crystal formation begins in most products
at a temperature of approximately 30° F and,
although some extremely concentrated fluids
still remain unfrozen even at temperatures below
—50° F, most of the fluids are solidified by the

* Air Conditioning Refrigerating Data Book,
Applications Volume, 5th Edition, American Society
of Refrigerating Engineers, 1954-55, p. 1-02.

time the product temperature is lowered to
25° F. The temperature range between 30° F
and 25° F is often referred to as the zone of
maximum ice-crystal formation, and rapid heat
removal through this zone is desirable from the
standpoint of product quality. This is particu-
larly true for fruits and vegetables because both
undergo serious tissue damage when slow frozen.
Since animal tissue is much tougher and
much more elastic than plant tissue, the freezing
rate is not as critical in the freezing of meats
and meat products as it is in fruits and vege-
tables. Recent experiment indicates that poultry
and fish suffer little, if any, cellular damage
when slow frozen. This does not mean, how-
ever, that quick frozen meats are not superior
to those which are slow frozen, but only that,
for the standpoint of cellular damage, quick
freezing is not as important in the freezing of
meats as it' is in fruits and vegetables. For
example, poultry that is slow frozen takes on a
darkened appearance which makes it much less
attractive to the consumer. This alone is
enough to justify the quick freezing of poultry.
Too, in all cases, quick freezing reduces the
processing time and, consequently, the amount

SURVEY OF REFRIGERATION APPLICATIONS 139

of bacterial deterioration. This is especially
worthwhile in the processing of fish because of
their tendency to rapid spoilage.
9-31. Packaging Materials. Dehydration, one
of the principal factors limiting the storage life
of frozen foods, is greatly reduced by proper
packaging. Unpackaged products are subject
to serious moisture losses not only during the
freezing process but also during the storage
period. While in storage, unpackaged frozen
products lose moisture to the air continuously
by sublimation. This eventually results in a
condition known as "freezer-burn," giving the
product a white, leathery appearance. Freezer-
burn is usually accompanied by oxidation,
flavor changes, and loss of vitamin content.

With few exceptions, all products are pack-
aged before being placed in frozen storage.
Although most products are packaged before
freezing, some, such as loose frozen peas and lima
beans, are packaged after the freezing process.

To provide adequate protection against
dehydration and oxidation, the packaging
material should be practically 100% gas and
vapor proof and should fit tightly around the
product to exclude as much air as possible. Too,
air spaces in packages have an insulating effect
which reduce the freezing rate and increase
freezing costs.

The fact that frozen products are in compe-
tition to products preserved by other methods
introduces several factors which must be taken
into account when selecting packaging materials.
When the product is to be sold directly to the
consumer, the package must be attractive and
convenient to use in order to stimulate sales.
From a cost standpoint, the package should be
relatively inexpensive and of such a nature that
it permits efficient handling so as to reduce
processing costs.

Some packaging materials in general use are
aluminum foil, tin cans, impregnated paper-
board cartons, paper-board cartons over-
wrapped with vapor-proof wrappers, wax paper,
cellophane, polyethylene, and other sheet
plastics.

Frozen fish are often given an ice glaze (a
thin coating of ice) which provides an excellent
protective covering. However, since the ice
glaze is very brittle, glazed fish must be handled
carefully to avoid breaking the glaze. Too, since
the ice glaze gradually sublimes to the air, the

fish must be reglazed approximately once a
month by dipping into fresh water or by
spraying.

9-32. Frozen Storage. The exact temperature
required for frozen storage is not critical,
provided that it is sufficiently low and that it
does not fluxuate. Although 0°F is usually
adequate for short-term (retail) storage, -5° F
is the best temperature for all-around long-term
(wholesale) storage. When products having
unstable fats (oxidizable, free, fatty acids) are
stored in any quantity, the storage temperature
should be held at —10° F or below in order to
realize the maximum storage life.

When products are stored above — 20° F,
which is normally the case, the temperature of
the storage room should be maintained constant
with a variation of not more than 1 ° F in either
direction. Variations in storage temperature
cause alternate thawing and refreezing of some
of the juices in the product. This tends to
increase the size of the ice crystals in the
product and eventually results in the same
type of cellular damage as occurs with slow
freezing.

Since many packaging materials do not offer
complete protection against dehydration, the
relative humidity should be kept at a high level
(85 % to 90%) in frozen storage rooms, particu-
larly for long-term storage.

Proper stacking of the product is also
essential. Stacking should always be such that
it permits adequate air circulation around the
product. It is particularly important to leave a
good size air space between the stored product
and the walls of the storage room. In addition
to permitting air circulation around the product,
this eliminates the possibility of the product
absorbing heat directly from the warm walls.
9-33. Commercial Refrigerators. The term
"commercial refrigerator" is usually applied to
the smaller, ready-built, refrigerated fixtures of
the type used by retail stores and markets,
hotels, restaurants, and institutions for the
processing, storing, displaying, and dispensing
of perishable commodities. The term is some-
times applied also to the larger, custom-built
refrigerated fixtures and rooms used for these
purposes.

Although there are a number of special
purpose refrigerated fixtures which defy classi-
fication, in general, commercial fixtures can

140 PRINCIPLES OF REFRIGERATION

Fig. 9-8. Typical reach-in refrigera-
tor. (Courtesy Tyler Refrigeration
Corporation.)

be grouped into three principal categories:
(1) reach-in refrigerators, (2) walk-in coolers,
and (3) display cases.

9-34. Reach-In Refrigerators. The reach-in
refrigerator is probably the most versatile and
the most widely used of all commercial fixtures.
Typical users are grocery stores, meat markets,
bakeries, drug stores, lunch counters, res-
taurants, florists, hotels, and institutions of all
kinds. Whereas some reach-in refrigerators
serve only a storage function, others are used
for both storage and display (Fig. 9-8). Those
serving only the storage function usually have
solid doors, whereas those used for display
have glazed doors.

9-35. Walk-In Coolers. Walk-in coolers are
primarily storage fixtures and are available in a
wide variety of sizes to fit every need. Nearly
all retail stores, markets, hotels, restaurants,
institutions, etc., of any size employ one or more
walk-in coolers for the storage of perishables
of all types. Some walk-in coolers are equipped
with glazed reach-in doors. This feature is
especially convenient for the storing, displaying,
and dispensing of such items as dairy products,
eggs, and beverages. Walk-in coolers with
reach-in doors are widely used in grocery stores,
particularly drive-in groceries, for handling
such items.

9-36. Display Cases. The principal function of
any kind of display fixture is to display the

product or commodity as attractively as possible
in order to stimulate sales. Therefore, in the
design of refrigerated display fixtures, first
consideration is given to the displaying of the
product. In many cases, this is not necessarily
compatible with providing the optimum storage
conditions for the product being displayed.
Hence, the storage life of a product in a display
fixture is frequently very limited, ranging from
a few hours in some instances to a week or more
in others, depending upon the type of product
and upon the type of fixture.

Display fixtures are of two general types:
(1) the self-service case, from which the customer
serves himself directly, and (2) the service case,
from which the customer is usually served by
an attendant. The former is very popular in
supermarkets and other large, retail, self-service
establishments, whereas the service case finds
use in the smaller groceries, markets, bakeries,
etc. Typical service cases are shown in Figs. 9-9
and 9-10.

Self-service cases are of two types, open and
closed, with the open type gaining rapidly in
popularity. With the advent of the super-
market, the trend has been increasingly toward
the open type self-service case, and the older,
closed type self-service cases are becoming
obsolete. Several of the more popular types of
open self-service cases are shown in Figs. 9-11
and 9-12. These are used to display meat

SURVEY OF REFRIGERATION APPLICATIONS 141

vegetables, fruit, frozen foods, ice cream, dairy
products, delicatessen items, etc. The design
of the case varies somewhat with the particular
type of product being displayed. Too, designs
are available for both wall and island installa-
tion. Although some provide additional storage
space, others do not.

9-37. Special Purpose Fixtures. Although all
the refrigerated fixtures discussed in the preced-
ing sections are available in a variety of designs
in order to satisfy the specific requirements of
individual products and applications, a number
of special purpose fixtures is manufactured
which may or may not fall into one of the three
general categories already mentioned. Some of
the more common special purpose fixtures are

Fig. 9-9. Conventional single-duty service case for
displaying meats. (Courtesy Tyler Refrigeration
Corporation.)

beverage coolers, milk coolers (dairy farm), milk
and beverage dispensers, soda fountains, ice
cream makers, water coolers, ice makers, back-
bar refrigerators, florist boxes, dough retarders,
candy cases, and mortuary refrigerators.
9-38. Frozen Food Locker Plants. Normally,
the function of a frozen food locker plant is to
process and freeze foods for individual families
and other groups, either for take-home storage
or for storage at the locker plant. When
storage is at the plant, the customer rents a
storage space (locker) and calls at the plant for
one or more packages as needed.

Fig. 9-10. Double-duty service case for displaying
meats. (Courtesy Tyler Refrigeration Corporation.)

As a general rule, a locker plant furnishes all
or most of the following facilities and/or
services:

1. A chilling room for chilling freshly killed
meats.

2. A cold storage room for holding products
under refrigeration while awaiting preparation
and processing prior to freezing.

3. A processing room where the products are
processed and packaged for the freezer.

Fig. 9-11. High multishelf produce sales
(Courtesy Tyler Refrigeration Corporation.)

142 PRINCIPLES OF REFRIGERATION

Fig. 9-12. Open-type display case for
frozen foods and icecream. (Courtesy
Tyler Refrigeration Corporation.)

4. A freezing room or cabinet in which the
food is frozen prior to being placed in storage.

5. A low temperature room containing the
storage lockers.

6. A low temperature bulk storage room.

7. An aging room where certain meats are
kept under refrigeration and allowed to age
(tenderize) for periods usually ranging from
7 to 10 days.

8. A curing and smoking room for handling
bacon, ham, sausage, and other cured meats.

Services such as slaughtering, lard rendering,
sausage making, etc., are also provided by some
plants.

The layout of a typical locker plant is shown
in Fig. 9-13. The recommended design condi-
tions for the various spaces in the locker plant
are given in Fig. 9-14. The average size of the

v//>y////Y//////^/Y//w//y/^/// , /yy/s/y/> , M^

Fig. 9-13. Typical locker plant. (ASRE Data Book, Applications Volume, 1956-57.) Reproduced by permission
of American Society of Heating, Refrigerating, and Air-Conditioning Engineers.

Type of space

SURVEY OF REFRIGERATION APPLICATIONS 143
Locker Plant Design Conditions

Refrigerant Insulation

Room
temperature

temperature thickness, inches.

Work room, process room,

and kitchen

Atmospheric

None

Chill room

34 to 36 F Design for

35 F

20 to 25 F below room
temperature, for gravity
circulation; 10 to 15 F
below room tempera-
ture for forced air cir-
culation

3 to 8

Aging room

34 to 36 F Design for

35 F

Same as chill room

3 to 8

Curing room

38 to 40 F Design for
40F

Same as chill room

3 to 8

Freezing room

(gravity

-10 to -20 F

-20 to -30 F

6 to 12

air circulation)

Freezer cabinet (in locker

Not important

-15 to -20 F

1 or 2

room)

Blast freezer

Depends on type of
system used

-10 to -15 F

6 to 12

Locker room or

bulk

-15 to -20 F

6 to 12

storage

Fig. 9-14. (ASRE Data Book, Applications Volume, 1956-57.
of Heating, Refrigerating and Air Conditioning Engineers.)

Reproduced by permission of American Society

individual locker is 6 cu ft and the average
product storage capacity is approximately 35
to 40 lb per cubic foot. Minimum product
turnover is approximately 2 lb per locker per
day. Standard practice is to base chilling room
and freezer capacities on the handling of 2 to 4
lb of product per locker per day.
9-39. Summary. Recognizing that a thorough
knowledge of the application itself is a pre-
requisite to good system design and proper
equipment selection, we have devoted the mate-
rial in this chapter to a brief survey of a few of
the applications of mechanical refrigeration,
with special emphasis being given to the area
of commercial refrigeration.

Obviously, the applications of mechanical
refrigeration are too many and too varied to
permit detailed consideration of each and
every type. Fortunately, this is neither necessary
nor desirable since methods of system designing
and equipment selection are practically the

same for all types of applications. Commerical
refrigeration was selected for emphasis because
this area embraces a wide range of applications
and because the problems encountered in this
area are representative of those in the other
areas. Hence, even though the discussion in
this chapter and in those which follow deals
chiefly with commercial refrigeration, the
principals of system design and the methods of
equipment selection developed therein may be
applied to all types of mechanical refrigeration
applications.

Although no attempt is made in this book
to discuss air conditioning as such except in a
very general way, it should be pointed out that
most commercial refrigeration applications,
particularly those concerned with product
storage, involve air conditioning in that they
ordinarily include close control of the tempera-
ture, humidity, motion, and cleanliness of the air
in the refrigerated space.

Cooling Load
Calculations

10-1. The Cooling Load. The cooling load on
refrigerating equipment seldom results from
any one single source of heat. Rather, it is the
summation of the heat which usually evolves
from several different sources. Some of the
more common sources of heat which supply
the load on refrigerating equipment are:

1. Heat that leaks into the refrigerated space
from the outside by conduction through the
insulated walls.

2. Heat that enters the space by direct
radiation through glass or other transparent
materials.

3. Heat that is brought into the space by
warm outside air entering the space through
open doors or through cracks around windows
and doors.

4. Heat given off by a warm product as its
temperature is lowered to the desired level.

5. Heat given off by people occupying the
refrigerated space.

6. Heat given off by any heat-producing
equipment located inside the space, such as
electric motors, lights, electronic equipment,
steam tables, coffee urns, hair driers, etc.

The importance of any one of these heat
sources with relation to the total cooling load
on the equipment varies with the individual
application. Not all them will be factors in
every application, nor will the cooling load in
any one application ordinarily include heat
from all these sources. However, in any given

application, it is essential that consideration be
given to all heat sources present and that all the
heat evolving from them be taken into account
in the over-all calculation.
10-2. Equipment Running Time. Although
refrigerating equipment capacities are normally
given in Btu per hour, in refrigeration applica-
tions the total cooling load is usually calculated
for a 24-hr period, that is, in Btu per 24 hr.
Then, to determine the required Btu per hour
capacity of the equipment, the total load for the
24-hr period is divided by the desired running
time for the equipment, viz:

RequiredBtu/hr Totalcoolingload)Btu/24hr

equipment = 2 : '.

capacity Desired running time

(10-1)

Because of the necessity for defrosting the
evaporator at frequent intervals, it is not
practical to design the refrigerating system in
such a way that the equipment must operate
continuously in order to handle the load. In
most cases, the air passing over the cooling coil
is chilled to a temperature below its dew point
and moisture is condensed out of the air onto
the surface of the cooling coil. When the tem-
perature of the coil surface is above the freezing
temperature of water, the moisture condensed
out of the air drains off the coil into the con-
densate pan and leaves the space through the con-
densate drain. However, when the temperature of
the cooling coil is below the freezing tempera-
ture of water, the moisture condensed out of
the air freezes into ice and adheres to the surface
of the coil, thereby causing "frost" to accumu-
late on the coil surface. Since frost accumulation
on the coil surface tends to insulate the coil
and reduce the coil's capacity, the frost must
be melted off periodically by raising the surface
temperature of the coil above the freezing point
of water and maintaining it at this level until
the frost has melted off the coil and left the
space through the condensate drain.

No matter how the defrosting is accomplished,
the defrosting requires a certain amount of
time, during which the refrigerating effect of the
system must be stopped.

One method of defrosting the coil is to stop
the compressor and allow the evaporator to
warm up to the space temperature and remain
at this temperature for a sufficient length of

144

COOLING LOAD CALCULATIONS 145

time to allow the frost accumulation to melt off
the coil. This method of defrosting is called
"off-cycle" defrosting. Since the heat required
to melt the frost in off-cycle defrosting must
come from the air in the refrigerated space,
defrosting occurs rather slowly and a consider-
able length of time is required to complete the
process. Experience has shown that when off-
cycle defrosting is used, the maximum allowable
running time for the equipment is 16 hr out of
each 24-hr period, the other 8 hr being allowed
for the defrosting. This means, of course, that
the refrigerating equipment must have sufficient
capacity to accomplish the equivalent of 24 hr of
cooling in 1 6 hr of actual running time. Hence,
when off-cycle defrosting is used, the equipment
running time used in Equation 10-1 is approxi-
mately 16 hr.

When the refrigerated space is to be main-
tained at a temperature below 34° F, off-cycle
defrosting is not practical. The variation in
space temperature which would be required in
order to allow the cooling coil to attain a
temperature sufficiently high to melt off the
frost during every off cycle would be detrimental
to the stored product. Therefore where the
space temperature is maintained below 34° F,
some method of automatic defrosting is
ordinarily used. In such cases the surface of the
coil is heated artificially, either with electric
heating elements, with water, or with hot gas
from the discharge of the compressor (see
Chapter 20).

Defrosting by any of these means is accom-
plished much more quickly than when off-cycle
defrosting is used. Hence, the off-cycle time
required is less for automatic defrosting and
the maximum allowable running time for the
equipment is greater than for the aforementioned
off-cycle defrosting. For systems using auto-
matic defrosting the maximum allowable
running time is from 18 to 20 hr out of each
24-hr period, depending upon how often de-
frosting is necessary for the application in
question. As a general rule, the 18 hr running
time is used.

It is of interest to note that since the tempera-
ture of the cooling coil in comfort air condition-
ing applications is normally around 40° F, no
frost accumulates on the coil surface and, there-
fore, no off-cycle time is required for defrosting.
For this reason, air conditioning systems are

usually designed for continuous run and cooling
loads for air conditioning applications are
determined directly in Btu per hour.
10-3. Cooling Load Calculations. To simplify
cooling load calculations, the total cooling load
is divided into a number of individual loads
according to the sources of heat supplying the
load. The summation of these individual loads
is the total cooling load on the equipment.

In commercial refrigeration, the total cooling
load is divided into four separate loads, viz:
(1) the wall gain load, (2) the air change load,
(3) the product load, and (4) the miscellaneous
or supplementary load.

10-4. The Wall Gain Load, The wall gain
load, sometimes called the wall leakage load,
is a measure of the heat which leaks through the
walls of the refrigerated space from the outside
to the inside. Since there is no perfect insulation,
there is always a certain amount of heat passing
from the outside to the inside whenever the
inside temperature is below that of the outside.
The wall gain load is common to all refrigeration
applications and is ordinarily a considerable
part of the total cooling load. Some exceptions
to this are liquid chilling applications, where the
outside area of the chiller is small and the walls
of the chiller are well insulated. In such cases,
the leakage of heat through the walls of the
chiller is so small in relation to the total cooling
load that its effect is negligible and it is usually
neglected. On the other hand, commercial
storage coolers and residential air conditioning
applications are both examples of applications
wherein the wall gain load usually accounts for
the greater portion of the total load.
10-5. The Air Change Load. When the door
of a refrigerated space is opened, warm outside
air enters the space to replace the more dense
cold air which is lost from the refrigerated space
through the open door. The heat which must
be removed from this warm outside air to
reduce its temperature to the space temperature
becomes a part of the total cooling load on the
equipment. This part of the total load is called
the air change load.

The relationship of the air change load to the
total cooling load varies with the application.
Whereas in some applications the air change
load is not a factor at all, in others it represents
a considerable portion of the total load. For
example, with liquid chillers, there are no doors

146 PRINCIPLES OF REFRIGERATION

or other openings through which air can pass
and therefore the air change load is nonexistent.
On the other hand, the reverse is true for air
conditioning applications, where, in addition
to the air changes brought about by door
openings, there is also considerable leakage of
air into the conditioned space through cracks
around windows and doors and in other parts
of the structure. Too, in many air conditioning
applications outside air is purposely introduced
into the conditioned space to meet ventilating
requirements. When large numbers of people
are in the conditioned space, the quantity of
fresh air which must be brought in from the
outside is quite large and the cooling load
resulting for the cooling of this air to the tem-
perature of the conditioned space is often a
large part of the total cooling load in such
applications.

In air conditioning applications, the air change
load is called either the ventilating load or the
infiltration load. The term ventilating load is
used when the air changes in the conditioned
space are the result of deliberate introduction
of outside air into the space for ventilating
purposes. The term infiltration load is used
when the air changes are the result of the
natural infiltration of air into the space through
cracks around windows and doors. Every air
conditioning application will involve either an
infiltration load or a ventilating load, but never
both in the same application.

Since the doors on commercial refrigerators
are equipped with well-fitted gaskets, the cracks
around the doors are tightly sealed and there is
little, if any, leakage of air around the doors of a
commercial fixture in good condition. Hence,
in commercial refrigeration, the air changes are
usually limited to those which are brought about
by actual opening and closing of the door or
doors.

10-6. The Product Load. The product load is
made up of the heat which must be removed
from the refrigerated product in order to reduce
the temperature of the product to the desired
level. The term product as used here is taken
to mean any material whose temperature is
reduced by the refrigerating equipment and
includes not only perishable commodities, such
as foodstuff, but also such items as welding
electrodes, masses of concrete, plastic, rubber,
and liquids of all kinds.

The importance of the product load in relation
to the total cooling load, like all others, varies
with the application. Although it is nonexistent
in some applications, in others it represents
practically the entire cooling load. Where the
refrigerated cooler is designed for product
storage, the product is usually chilled to the
storage temperature before being placed in the
cooler and no product load need be considered
since the product is already at the storage
temperature. However, in any instance where
the product enters a storage cooler at a tem-
perature above the storage temperature, the
quantity of heat which must be removed from
the product in order to reduce its temperature
to the storage temperature must be considered
as a part of the total load on the cooling
equipment.

In some few instances, the product enters the
storage fixture at a temperature below the
normal storage temperature for the product.
A case in point is ice cream which is frequently
chilled to a temperature of 0°F or -10° F
during the hardening process, but is usually
stored at about 10° F, which is the ideal dipping
temperature. When such a product enters
storage at a temperature below the space
temperature, it will absorb heat from the
storage space as it warms up to the storage
temperature and thereby produce a certain
amount of refrigerating effect of its own. In
other words, it provides what might be termed
a negative product load which could theoreti-
cally be subtracted from the total cooling load.
This is never done, however, since the refriger-
ating effect produced is small and is not
continuous in nature.

The cooling load on the refrigerating equip-
ment resulting from product cooling may be
either intermittent or continuous, depending
on the application. The product load is a part
of the total cooling load only while the tempera-
ture of the product is being reduced to the
storage temperature. Once the product is
cooled to the storage temperature, it is no
longer a source of heat and the product load
ceases to be a part of the load on the equipment.
An exception to this is in the storage of fruit
and vegetables which give off respiration heat
for the entire time they are in storage even
though there is no further decrease in their
temperature (see Section 10-17).

COOLING LOAD CALCULATIONS 147

There are, of course, a number of refrigera-
tion applications where product cooling is
more or less continuous, in which case the
product load is a continuous load on the equip-
ment. This is true, for instance, in chilling
coolers where the primary function is to chill
the warm product to the desired storage tem-
perature. When the product has been cooled
to the storage temperature, it is usually moved
out of the chilling room into a storage room and
the chilling room is then reloaded with warm
product. In such cases, the product load is
continuous and is usually a large part of the
total load on the equipment.

Liquid chilling is another application wherein
the product provides a continuous load on the
refrigerating equipment. The flow of the liquid
being chilled through the chiller is continuous
with warm liquid entering the chiller and cold
liquid leaving. In this instance, the product
load is practically the only load on the equip-
ment since there is no air change load and the
wall gain load is negligible, as is the miscel-
laneous load.

In air conditioning applications there is no
product load as such, although there is often a
"pull-down load," which, in a sense, may be
thought of as a product load.
10-7. The Miscellaneous Load. The miscel-
laneous load, sometimes referred to as the
supplementary load, takes into account all
miscellaneous sources of heat. Chief among
these are people working in or otherwise
occupying the refrigerated space along with
lights or other electrical equipment operating
inside the space.

In most commercial refrigeration applica-
tions the miscellaneous load is relatively small,
usually consisting only of the heat given off by
lights and fan motors used inside the space.

In air conditioning applications, there is no
miscellaneous load as such. This is not to say
that human occupancy and equipment are not
a part of the cooling load in air conditioning
applications. On the contrary, people and
equipment are often such large factors in the
air conditioning load that they are considered
as separate loads and are calculated as such.
For example, in those air conditioning applica-
tions where large numbers of people occupy the
conditioned space, such as churches, theaters,
restaurants, etc., the cooling load resulting from

human occupancy is frequently the largest single
factor in the total load. Too, many air condi-
tioning systems are installed for the sole purpose
of cooling electrical, electronic, and other types
of heat-producing equipment. In such cases,
the equipment usually supplies the greater
portion of the cooling load.
10-8. Factors Determining the Wall Gain
Load. The quantity of heat transmitted through
the walls of a refrigerated space per unit of time
is the function of three factors whose relation-
ship is expressed in the following equation:

Q=A x U x D

(10-2)

where Q = the quantity of heat transferred in
Btu/hr
A — the outside surface area of the wall

(square feet)
U = the over-all coefficient of heat trans-
mission (Btu/hr/sq ft/° F)
D = the temperature differential across
the wall (° F)

The coefficient of transmission or "U" factor
is a measure of the rate at which heat will pass
through a 1 sq ft area of wall surface from the
air on one side to the air on the other side for
each 1° F of temperature difference across the
wall. The value of the U factor is given in
Btu per hour and depends on the thickness of
the wall and on the materials used in the wall
construction. Since it is desirable to prevent as
much heat as possible from entering the space
and becoming a load on the cooling equipment,
the materials used in the construction of cold
storage walls should be good thermal insulators
so that the value of U is kept as low as is
practical.

According to Equation 10-2, once the U
factor is established for a wall, the rate of heat
flow through the wall varies directly with the
surface area of the wall and with the temperature
differential across the wall. Since the value of
U is given in Btu/hr/sq ft/° F, the total quantity
of heat passing through any given wall in 1 hr
can be determined by multiplying the U factor
by the wall area in square feet and by the tem-
perature difference across the wall in degrees
Fahrenheit, that is, by application of Equation
10-2.

Example 10-1. Determine the total quan-
tity of heat in Btu per hour which will pass
through a wall 10 ft by 20 ft, if the U factor

148 PRINCIPLES OF REFRIGERATION

for the wall is 0.16 Btu/hr/sq ft/ F and the
temperature on one side of the wall is 40° F
while the temperature on the other side is 95° F.

Solution
Total wall area

Temperature differ-
ential across wall, ° F

Applying Equation
10-2, the heat gain
through the wall

■ 10 ft x 20 ft
= 200 sq ft

. 95° - 40°
>55°F

200 x 0.16 x 55
1760 Btu/hr

Since the value of U in Equation 10-2 is in
Btu per hour, the result obtained from Equation
10-2 is in Btu per hour. To determine the wall
gain load in Btu per 24 hr as required in refriger-
ation load calculations, the result of Equation
10-2 is multiplied by 24 hr. Hence, for calcula-
tion cooling loads in refrigeration applications,
Equation 10-2 is written to include this multi-
plication, viz:

Q=AxUxDx24

(10-3)

10-9. Determination of the (/Factor. Over-
all coefficients of transmission or U factors have
been determined for various types of wall
construction and these values are available in
tabular form. Tables 10-1 through 10-3 list U
values for various types of cold storage walls.

Example 10-2. From Table 10-1, determine
the U factor for a wall constructed of 6-in. clay
tile with 4 in. of corkboard insulation.

Solution. Turn to Table 10-1 and select the
appropriate type of wall construction (third

Air spaces

Concrete
aggregate

Fig. 10-1. Concrete aggregate building block.

from top). In the next column select the desired
thickness of clay tile (6 in.) and move to the
right to the column listing values for 4 in. of
insulation. Read the U factor of the wall, 0.064
Btu/hr/sq ft/° F.

Should it be necessary, the U factor for any
type of wall construction can be readily calcu-
lated provided that either the conductivity or the
conductance of each of the materials used in the
wall construction is known. The conductivity
or conductance of most of the materials used in
wall construction can be found in tables. Too,
this information is usually available from the
manufacturer or producer of the material.
Table 10-4 lists the thermal conductivity or the
conductance of materials frequently used in the
construction of cold storage walls.

The thermal conductivity or k factor of a
material is the rate in Btu per hour at which
heat passes through a 1 sq ft cross section of the
material 1 in. thick for each 1 ° F of temperature
difference across the material.

Whereas the thermal conductivity or k factor
is available only for homogeneous materials
and the value given is always for a 1 in. thickness
of the material, the thermal conductance or C
factor is available for both homogeneous and
nonhomogeneous materials and the value given
is for the specified thickness of the material.

For any homogeneous material, the thermal
conductance can be determined for any given
thickness of the material by dividing the k
factor by the thickness in inches. Hence, for a
homogeneous material,

C=\ (10-4)

where x = the thickness of material in inches.

Example 10-3. Determine the thermal
conductance for a 5 in. thickness of corkboard.

Solution
From Table 10-4,
k factor of cork-
board = 0.30 Btu/hr/sq ft/in/° F

Applying Q ,

Equation 10-4, C =_

= 0.06 Btu/hr/sq ft/° F

Since the rate of heat transmission through
nonhomogeneous materials, such as the concrete
building block in Fig. 10-1, will vary in the

COOLING LOAD CALCULATIONS 149

several parts of the material, the C factor from
nonhomogeneous materials must be determined
by experiment.

The resistance that a wall or a material offers
to the flow of heat is inversely proportional to
the ability of the wall or material to transmit
heat. Hence, the over-all thermal resistance of
a wall can be expressed as the reciprocal of the
over-all coefficient of transmission, whereas the
thermal resistance of an individual material
can be expressed as the reciprocal of its conduc-
tivity or conductance, viz:

Over-all thermal resist-
ance

Thermal resistance of
an individual material

1 1 x

-l or c or k

The terms I Ik and 1/C express the resistance
to heat flow through a single material from
surface to surface only and do not take into
account the thermal resistance of the thin film
of air which adheres to all exposed surfaces. In
determining the over-all thermal resistance to
the flow of heat through a wall from the air on
one side to the air on the other side, the resist-
ance of the air on both sides of the wall should
be considered. Air film coefficients or surface
conductances for average wind velocities are
given in Table 10-5/*.

When a wall is constructed of several layers
of different materials the total thermal resistance
of the wall is the sum of the resistances of the
individual materials in the wall construction,
including the air films, viz:

\ \ x x x \

U fi **1 *2 k n Jo

Therefore
U *

I X X x 1

ft k x k t k n f„

where rr = surface conductance of inside wall,

' ' floor, or ceiling

— = surface conductance of outside wall,

■* * floor, or roof

Note. When nonhomogeneous materials are
used, 1/C is substituted for xjk.

Example 10-4. Calculate the value of £/for
a wall constructed of 8 in. cinder aggregate
building blocks, insulated with 4 in. of cork-
board, and finished on the inside with 0.5 in. of
cement plaster.

Solution
From Table 10-4,

8 in. cinder

aggregate block

C=0.60

Corkboard

k = 0.30

Cement plaster

k =8.00

From Table

10-5,4,

inside

surface

conductance

ft = 1-65

outside

surface

conductance

/„ = 4.00

Applying

Equation 10-5,

-I+-L+-L

the over-all

4 0.6 0.3

thermal resist-

. 0.5 1

ance, 1/17

+ T + L65

= 0.25 + 1.667 + 13.333

+ 0.0625 + 0.607

= 15.92

Therefore, U

= 1/15.92

= 0.0622 Btu/hr/sq ft/ F

For the most part, it is the insulating material
used in the wall construction that determines
the value of U for cold storage walls. The
surface conductances and the conductances of
the other materials in the wall have very little
effect on the value of U because the thermal
resistance of the insulating material is so large
with relation to that of the air films and other
materials. Therefore, for small coolers, it is
sufficiently accurate to use the conductance of
the insulating material alone as the wall U
factor.

10-10. Temperature Differential across Cold
Storage Walls. The temperature differential
across cold storage walls is usually taken as the
difference between the inside and outside design
temperatures.

The inside design temperature is that which
is to be maintained inside the refrigerated space
and usually depends upon the type of product
to be stored and the length of time the product
is to be kept in storage. The recommended
storage temperatures for various products are
given in Tables 10-10 through 10-13.

150 PRINCIPLES OF REFRIGERATION

The outside design temperature depends on
the location of the cooler. For cold storage
walls located inside a building, the outside
design temperature for the cooler wall is taken
as the inside temperature of the building. When
cold storage walls are exposed to the outdoors,
the outdoor design temperature for the region
(Table 10-6) is used as the outside design tem-
perature. The outdoor design temperatures
given in Table 10-6 are average outdoor tem-
peratures and include an allowance for normal
variations in the outdoor design dry bulb
temperature during a 24-hr period. These
temperatures should not be used for calculating
air conditioning loads.

1 0-1 1.TemperatureDifferential across Ceil-
ings and Floors. When a cooler is located
inside of a building and there is adequate
clearance between the top of the cooler and the
ceiling of the building to allow free circulation of
air over the top of the cooler, the ceiling of the
cooler is treated the same as an inside wall.
Likewise, when the top of the cooler is exposed
to the outdoors, the ceiling is treated as an
outdoor wall. The same holds true for floors
except when the cooler floor is laid directly on a
slab on the ground. As a general rule, the
ground temperature under a slab varies only
slightly the year round and is always consider-
ably less than the outdoor design dry bulb
temperature for the region in summer. Ground
temperatures used in determining the tempera-
ture differential across the floor of cold storage
rooms are given in Table 10-6A and are based
on the regional outdoor design dry bulb
temperature for winter.

10-12. Effect of Solar Radiation. Whenever
the walls of a refrigerator are so situated that
they receive an excessive amount of heat by
radiation, either from the sun or from some
other hot body, the outside surface temperature
of the wall will usually be considerably above
the temperature of the ambient air. A familiar
example of this phenomenon is the excessive
surface temperature of an automobile parked
in the sun. The temperature of the metal
surface is much higher than that of the sur-
rounding air. The amount by which the surface
temperature exceeds the surrounding air tem-
perature depends upon the amount of radiant
energy striking the surface and upon the
reflectivity of the surface. Recall (Section 2-21)

that radiant energy waves are either reflected
by or absorbed by any opaque material that
they strike. Light-colored, smooth surfaces
will tend to reflect mbre and absorb less radiant
energy than dark, rough-textured surfaces.
Hence, the surface temperature of smooth,
light-colored walls will be somewhat lower than
that of dark, rough-textured walls under the
same conditions of solar radiation.

Since any increase in the outside surface
temperature will increase the temperature
differential across the wall, the temperature
differential across sunlit walls must be corrected
to compensate for the sun effect. Correction
factors for sunlit walls are given in Table 10-7.
These values are added to the normal tempera-
ture differential. For walls facing at angles to
the directions listed in Table 10-7, average
values can be used.

10-13. Calculating the Wall Gain Load. In
determining the wall gain load, the heat gain
through all the walls, including the floor and
ceiling, must be taken into account. When
the several walls or parts of walls are of different
construction and have different U factors, the
heat leakage through the different parts is
computed separately. Walls having identical
U factors may be considered together, provided
that the temperature differential across the
walls is the same. Too, where the difference
in the value of U is slight and/or the wall area
involved is small, the difference in the U factor
can be ignored and the walls or parts of walls
can be grouped together for computation.

Example 10-5. A walk-in cooler, 16 ft x
20 ft x 10 ft high is located in the southwest
corner of a store building in Dallas, Texas (Fig.
10-2). The south and west walls of the cooler
are adjacent to and a part of the south and west
walls of the store building. The store has a 14 ft
ceiling so that there is a 4 ft clearance between
the top of the cooler and the ceiling of the store.
The store is air conditioned and the temperature
inside the store is maintained at approximately
80° F. The inside design temperature for the
cooler is 35° F. Determine the wall gain load
for the cooler if the walls of the cooler are of the
following construction:

South and west
(outside walls) 6 in. clay tile
6 in. corkboard
0.S cement plaster finish
on inside

COOLING LOAD CALCULATIONS

151

North and east
(inside walls)

Ceiling

Floor

Solution

Wall surface area

North wall

West wall

South wall

East wall

Ceiling

Floor

1 in. board on both sides

of 2 x 4 studs
3f in. granulated cork
Same as north and east

walls
4 in. corkboard laid on

5 in. slab and finished

with 3 in. of concrete

10 x 16 = 160 sq ft
10 x 20 = 200 sq ft
10 x 16 = 160sqft
10 x 20 = 200 sq ft
16 x 20 = 320 sq ft
16 x 20 = 320 sq ft
Wall U factors (Tables 10-1, 10-2, and 10-3)

North and east

walls
South and west

walls
Ceiling
Floor
From Table 10-6,
outside summer de-
sign dry bulb for
Dallas

From Table
10-6A, design
ground temperature
for Dallas

0.079 Btu/hr/sq ft/° F

0.045
0.079
0.066

92° F

70° F

20' :

, Outside design

16 ' H temperature, 92* F

M I MM I MIM ■ ■

6 in. corkboard-' | 6 in. clay tile

Cooler 35° F
10 ft ceiling

Partitions 3| in.

granulated cork-
1 in. board
on each side

^ mm

Inside temperature

80' F

Ceiling 14 ft

Fig. 10-2

A short method calculation may be used to
determine the wall gain load for small coolers
and for large coolers where the wall U factor

Outside
Design
Temp.

Inside
Design
Temp.

Normal
Wall
T.D.

Correction

Factor from

Table 10-7

Design
Wall
T.D.

North wall
South wall
West wall
East wall
Ceiling
Floor

80° F
92° F
92° F
80° F
80° F
70° F

35° F
35° F
35* F
35° F
35° F
35° F

45° F
57° F
57= p
45° F
45° F
35° F

4°F

6°F

45° F
61° F
63° F
45° F
45° F
35° F

Applying Equation 10-2,

North wall 160 x 0.079 x 45 = 569Btu/hr

West wall 200 x 0.045 x 63 = 567

South wall 160 x 0.045 x 61 = 439

East wall 200 x 0.079 x 45 = 7ll

Ceiling 320 x 0.079 x 45 = 1,137

Floor 320 x 0.066 x 35 = 739

Total wall gain load 4 . 162 Btu/hr

= 4,162 x 24 - 99,890 Btu/24 hr

and temperature difference are approximately
the same for all the walls. Table 10-18 lists wall
gain factors (Btu/24 hr sq ft) based on the thick-
ness of the wall insulation and on the tempera-
ture differential across the wall. To compute
the wall gain load in Btu/24 hr by the short
method, multiply the total outside wall area
(including floor and ceiling) by the appropriate
wall gain factor from Table 10-18, viz:

152 PRINCIPLES OF REFRIGERATION

Wall gain load = Outside surface area

x wall gain factor

To select the appropriate wall gain factor
from Table 10-18, find the thickness of the wall
insulation in the extreme left-hand column of
the table, move right to the column headed by
the design wall temperature difference, and read
the wall gain factor in Btu/24 hr/sq ft. For
example, assume that the walls of a cooler are
insulated with the equivalent of 4 in. of cork-
board and that the temperature difference
across the walls is 55° F. From Table 10-18,
read the wall gain factor of 99 Btu/24 hr/sq ft
(see Example 10-18).

10-14. Calculation the Air Change Load.
The space heat gain resulting from air changes
in the refrigerated space is difficult to determine
with any real accuracy except in those few cases
where a known quantity of air is introduced
into the space for ventilating purposes. When
the weight of outside air entering the space in a
24-hr period is known, the space heat gain
resulting from air changes depends upon the
difference in the enthalpy of the air at the
inside and outside conditions and can be
calculated by applying the following equation:

Air change load = W(h„ - hi) (10-6)

where W = weight of air entering space in 24 hr
(lb/24 hr)
h, = enthalpy of outside air (Btu/lb)
hi = enthalpy of inside air (Btu/lb)

However, since air quantities are usually
given in cubic feet rather than in pounds, to
facilitate calculations the heat gain per cubic
foot of outside air entering the space is listed in
Tables 10-8A and 10-8B for various inside and
outside air conditions. To determine the air
change load in Btu per 24 hr, multiply the air
quantity in cubic feet per 24 hr by the appropri-
ate factor from Table 10-8 A or 10-8B.

Where the ventilating air (air change) quantity
is given in cubic feet per minute (cfm), convert
cfm to cubic feet per 24 hr by multiplying by
60 min and by 24 hr.

Example 10-6. Three hundred cfm of air
are introduced into a refrigerated space for
ventilation. If the inside of the cooler is main-
tained at 35° F and the outside dry bulb tem-
perature and humidity are 85° F and 50%,
respectively, determine the air change load in
Btu/24 hr.

Solution
Cubic feet of air per
per24hr

From Table 10-8.A
heat gain per cubic
feet

Ventilating (air
change) load

cfm x 60 x 24
300 x 60 x 24
432,000 cu ft/24 hr

= 1.86Btu/cuft

= cu ft/24 hr
x Btu/cu ft
= 432,000 x 1.86
= 803,520 Btu/24 hr

Except in those few cases where air is pur-
posely introduced into the refrigerated space for
ventilation, the air changes occurring in the
space are brought about solely by infiltration
through door openings, The quantity of outside
air entering a space through door openings in
a 24-hr period depends upon the number, size,
and location of the door or doors, and upon
the frequency and duration of the door open-
ings. Since the combined effect of all these
factors varies with the individual installation
and is difficult to predict with reasonable ac-
curacy, it is general practice to estimate the
air change quantity on the basis of experience
with similar applications. Experience has shown
that, as a general rule, the frequency and dur-
ation of door openings and, hence, the air
change quantity, depend on the inside volume
of the cooler and the type of usage. Tables
10-9A and 10-9B list the approximate number
of air changes per 24 hr for various cooler
sizes. The values given are for average usage
(see table footnotes). The ASRE Data Book
defines average and heavy usage as follows:

Average usage includes installations not
subject to extreme temperatures and where
the quantity of food handled in the refrigerator
is not abnormal. Refrigerators in delicatessens
and clubs may generally be classified under this
type of usage.

Heavy usage includes installations such as
those in busy markets, restaurant and hotel
kitchens where the room temperatures are likely
to be high, where rush periods place heavy loads
on the refrigerator, and where large quantities
of warm foods are often placed in it.*

* The Refrigerating Data Book, Basic Volume,
The American Society of Refrigerating Engineers,
1949, New York, p. 327.

Example 10-7. A walk-in cooler 8 ft x
IS ft x 10 ft high is constructed of 4 in. of cork-
board with 1 in. of wood on each side. The
outside temperature is 95° F and the humidity
is 50%. The cooler is maintained at 35° F and
the usage is average. Determine the air change
loadinBtu/24hr.

Solution. Since the walls of the cooler are
approximately 6 in. thick (4 in. of corkboard
and 2 in. of wood), the inside dimensions of the
cooler are 1 ft less than the outside dimensions;
therefore,
Inside volume = 7 ft x 14 ft x 9 ft

= 882 cu ft
From Table 10-9A,
by interpolation, num-
ber of air changes per
24 hr for cooler vol-
ume of approximately
900cuft =19

Total quantity of
air change per 24 hr = Inside volume

x air changes
= 882 x 19
= 16,758 cu ft/24 hr
From Table 10-8A,
heat gain per cubic
feet = 2.49 Btu/cu ft

Air change load = cu ft/24 hr

x Btu/cu ft
= 16,758 x 2.49
= 41,727 Btu/24 hr

10-15. Calculation the Product Load. When
a product enters a storage space at a tempera-
ture above the temperature of the space, the
product will give off heat to the space until it
cools to the space temperature. When the
temperature of the storage space is maintained
above the freezing temperature of the product,
the amount of heat given off by the product in
cooling to the space temperature depends Upon
the temperature of the space and upon the
weight, specific heat, and entering temperature
of the product. In such cases, the space heat
gain from the product is computed by the
following equation, (see Section 2-24):

Q = W x C x(T a -Tj (10-7)
where Q = the quantity of heat in Btu

W = weight of the product (pounds)
C = the specific heat above freezing

(Btu/lb/° F)
T x = the entering temperature (° F)
T a = the space temperature (° F)

COOLING LOAD CALCULATIONS 153

Example 10-8. One thousand pounds of
fresh, lean beef enter a cooler at 55° F and are
chilled to the cooler temperature of 35° F in
24 hr. Calculate the product load in Btu/24 hr.

Solution
From Table 10-12, the
specific heat of lean
beef above freezing = 0.75 Btu/lb/° F

Applying Equation
10-7, the product load,
Btu/24 hr = 1000 x 0.75

x (55 - 35)
= 1000 x 0.75 x 20
= 15,000 Btu/24 hr

Notice that no time element is inherent in
Equation 10-7 and that the result obtained is
merely the quantity of heat the product will
give off in cooling to the space temperature.
However, since in Example 10-8 the product is
to be cooled over a 24-hr period, the resulting
heat quantity represents the product load for a
24-hr period. When the desired cooling time
is less than 24 hr, the equivalent product load
for a 24-hr period is computed by dividing the
heat quantity by the desired cooling time for the
product to obtain the hourly cooling rate and
then multiplying the result by 24 hr to determine
the equivalent product load for a 24-hr period.
When adjusted to include these two factors,
Equation 10-7 is written:

WxCx(r,-r 1 )x24hr

Q aav J n*s 0°- 8 )

desired cooling time (hr)

Example 104. Assume that it is desired to
chill the beef in Example 10-8 in 6 hr rather than
in 24 hr. Determine the product load in
Btu/24 hr.

Solution. Applying
Equation 10-8, product
load, Btu/24 hr

1000 x 0.75
_ x (55 - 35) x 24

6
= 60,000 Btu/24 hr

Compare this result with that obtained in
Example 10-8.

When a product is chilled and stored below
its freezing temperature, the product load is
calculated in three parts:

1. The heat given off by the product in cooling
from the entering temperature to its freezing
temperature.

154 PRINCIPLES OF REFRIGERATION

2. The heat given off by the product in solidi-
fying or freezing.

3. The heat given off by the product in cool-
ing from its freezing temperature to the final
storage temperature.

For parts 1 and 3, Equation 10-7 is used to
determine the heat quantity. For part 1, T x in
Equation 10-7 is the entering temperature of the
product, whereas T z is the freezing temperature
of the product (Tables 10-10 through 10-13).
For part 3, T x in Equation 10-7 is the freezing
temperature of the product and T t is the final
storage temperature. The heat quantity for
part 2 is determined by the following equation:

Q = W x h if (10-9)

where W = the weight of the product (pounds)
h it = the latent heat of the product
(Btu/lb)

When the chilling and freezing are accom-
plished over a 24-hr period, the summation of
the three parts represents the product load for
24 hr. When the desired chilling and freezing
time for the product are less than 24 hr, the
summation of the three parts is divided by the
desired processing time and then multiplied by
24 hr to determine the equivalent 24-hr product
load.

Example 10-10. 500 pounds of poultry
enter a chiller at 40° F and are frozen and chilled
to a final temperature of — 5° F for storage in
12 hr. Compute the product load in Btu/24 hr.

Solution
From Table 10-12,
Specific heat above

freezing
Specific heat below

freezing
Latent heat
Freezing

temperature

To cool poultry from
entering temperature
to freezing tempera-
ture, applying Equa-
tion 10-7

= 0.79 Btu/lb/° F

= 0.37 Btu/lb/° F
= 106 Btu/lb

= 27°F

To freeze, applying
Equation 10-9

500 x 0.79
x (40 - 27)
5135 Btu

500 x 106
53,000 Btu

To cool from freezing
temperature to final
storage temperature,
applying Equation
10-7

500 x 0.37
x[27-(-5)]

5920 Btu

Total heat given up
by product (summa-
tion of 1,2, and 3)

Equivalent product
load for 24-hrperiod
Btu/24 hr

= 64,000 Btu

64,000 x 24 hr

12 hr

= 128,000 Btu/24 hr

10-16. Chilling Rate Factor. During the early
part of the chilling period, the Btu per hour load
on the equipment is considerably greater than the
average hourly product load as calculated in
the previous examples. Because of the high
temperature difference which exists - between the
product and the space air at the start of the
chilling period, the chilling rate is higher and
the product load tends to concentrate in the
early part of the chilling period (Section 9-23).
Therefore, where the equipment selection is
based on the assumption that the product load
is evenly distributed over the entire chilling
period, me equipment selected will usually have
insufficient capacity to carry the load during
the initial stages of chilling when the product
load is at a peak.

To compensate for the uneven distribution of
the chilling load, a chilling rate factor is intro-
duced into the chilling load calculation. The
effect of the chilling rate factor is to increase the
product load calculation by an amount sufficient
to make the average hourly cooling rate approxi-
mately equal to the hourly load at the peak
condition. This results in the selection of
larger equipment, having sufficient capacity to
carry the load during the initial stages of chilling.

Chilling rate factors for various products are
listed in Tables 10-10 through 10-13. The
factors given in the tables are based on actual
tests and on calculations and will vary with the
ratio of the loading time to total chilling time.
As an example, test results show that in typical
beef and hog chilling operations the chilling rate
is 50% greater during the first half of the chill-
ing period than the average chilling rate for
the entire period. The calculation without the
chilling rate factor will, of course, show the

average chilling rate for the entire period. To
obtain this rate during the initial chilling period,
it muSt be multiplied by l.S. For convenience,
the chilling rate factors are given in the tables
in reciprocal form and are used in the denomi-
nator of the equation. Thus the chilling rate
factor for beef as shown in the table is 0.67
0/1.5).

Where a chilling rate factor is used, Equation
10-7 is written

W xC xjTi-TJ
Q ~ Chilling rate factor (1 °" 10)

As a general rule chilling rate factors are not
used for the final stages of chilling from the
freezing temperature to the final storage tem-
perature of the product. Too, chilling rate
factors are usually applied to chilling rooms
only and are not normally used in calculation
of the product load for storage rooms. Since
the product load for storage rooms usually
represents only a small percentage of the total
load, the uneven distribution of the product
load over the cooling period will not ordinarily
cause overloading of the equipment and, there-
fore, no allowance need be made for this
condition.

10-17. Respiration Heat. Fruits and vegetables
are still alive after harvesting and continue to
undergo changes while in storage. The more
important of these changes are produced by
respiration, a process during which oxygen
from the air combines with the carbohydrates
in the plant tissue and results in the release of
carbon dioxide and heat. The heat released is
called respiration heat and must be considered
as a part of the product load where considerable
quantities of fruit and/or vegetables are held in
storage. The amount of heat evolving from the
respiration process depends upon the type and
temperature of the product. Respiration heat
for various fruits and vegetables is listed in
Table 10-14.

Since respiration heat is given in Btu per
pound per hr, the product load accruing from
respiration heat is computed by multiplying the
total weight of the product by the respiration
heat as given in Table 10-14, viz:

Q (Btu/24 hr) - Weight of product (lb)

x respiration heat (Btu/lb/hr)
x24hr (10-11)

COOLING LOAD CALCULATIONS 155

10-18. Containers and Packing Materials.

When a product is chilled in containers, such as
milk in bottles or cartons, eggs in crates, fruit
and vegetables in baskets and lugs, etc., the
heat given off by the containers and packing
materials in cooling from the entering tempera-
ture to the 'space temperature must be con-
sidered as a part of the product load. Equation
10-7 is used to compute this heat quantity.
10-19. Calculating the Miscellaneous Load.
the miscellaneous load consists primarily of the
heat given off by lights and electric motors
operating in the space and by people working
in the space. The heat given off by lights is
3.42 Btu per watt per hour. The heat given off
by electric motors and by people working in the
space is listed in Tables 10-15 and 10-16,
respectively. The following calculations are
made to determine the heat gain from mis-
cellaneous:

Lights: wattage x 3.42 Btu/watt/hr x 24 hr

Electric motors : factor (Table 10-1 5) x horse-
power x 24 hr

People: factor (Table 10-16) x number of
people x 24 hr

10-20. Use of Safety Factor. The total cooling
load for a 24-hr period is the summation of the
heat gains as calculated in the foregoing sections.
It is common practice to add 5% to 10% to
this value as a safety factor. The percentage
used depends upon the reliability of the informa-
tion used in calculating the cooling load. As a
general rule 10% is used.

After the safety factor has been added, the
24-hr load is divided by the desired operating
time for the equipment to determine the average
load in Btu per hour (see Section 10-2). The
average hourly load is used as a basis for
equipment selection.

10-21. Short Method Load Calculations.
Whenever possible the cooling load should be
determined by using the procedures set forth
in the preceding sections of this chapter. How-
ever, when small coolers (under 1600 cu ft) are
used for general-purpose storage, the product
load is frequently unknown and/or varies
somewhat from day to day so that it is not pos-
sible to compute the product load with any real
accuracy. In such cases, a short method of load
calculation can be employed which involves the
use of load factors which have been determined
by experience. When the short method of

156 PRINCIPLES OF REFRIGERATION

calculation is employed, the entire cooling load
is divided into two parts: (1) the wall gain load
and (2) the usage or service load.

The wall gain load is calculated as outlined
in Section 10-13. The usage load is computed
by the following equation:

Usage load = interior volume x usage factor

(10-12)

Notice that the usage factors listed in Table
10-17 vary with the interior volume of the cooler
and with the difference in temperature between
the inside and outside of the cooler. Too, an
allowance is made for normal and heavy usage.
Normal and heavy usage have already been
defined in Section 10-14. No safety factor is
used when the cooling load is calculated by
the short method. The total cooling load is
divided by the desired operating time for the
equipment to find the average hourly load used
to select the equipment.

Example 10-11. A walk-in cooler 18 ft x
10 ft x 10 ft high has 4 in. of corkboard insula-
tion and standard wall construction consisting
of two layers of paper and 1 in. of wood on each
side (total wall thickness is 6 in.). The tempera-
ture outside the cooler is 85° F. 35001b of
mixed vegetables are cooled 10° F to the storage
temperature each day. Compute the cooling
load in Btu/hr based on a 16-hr per day opera-
ting time for the equipment. The inside
temperature is 40° F.

Solution
Outside surface area = 2 x 18 ft x 10 ft

- 360 sq ft
= 4 x 10 ft x 10 ft
= 400 sq ft
760 sq ft
Inside volume (since
total wall thickness is
6 in., the inside dimen-
sions are 1 ft less than
the outside dimensions)

Wall gain factor
(Table 10-18) (45° F
TD and 4 in. insulation)

Air changes (Table
10-9A) (by interpola-
tion)

Heat gain per cubic
foot of air (Table
10-8A)(50%RH)

17ft x9ft x9ft
1377 cu ft

81 Btu/hr/sq ft
■■ 16.8 per 24 hr
• 1.69 Btu/cuft

Average specific heat
of vegetables (Table
10-11) = 0.9 Btu/lb/° F

Average respiration
heat of vegetables
(Table 10-14) = 0.09 Btu/lb/hr

Wall gain load:
Area x wall gain factor
= 760 sq ft x 81 Btu/sq ft/24 hr

= 61,560 Btu/24 hr
Air change load:
Inside volume

x air changes x Btu/cuft
= 1377cuft x 16.8
x 1.69 Btu/cuft = 23,100 Btu/24 hr
Product load:
Temperature reduction

= Mxcx(r,-r 1 )

= 35001b x0.9Btu/lb/°F

x 10° F = 31,500 Btu/24 hr

Respiration heat
= M x reaction heat x 24 hr
= 3500 x 0.09 Btu/lb/hr
x24hr

7,560 Btu/24 hr
Summation:
Safety factor

(10%) =1 2,370 Btu

Total cooling

load = 136,100 Btu/24 hr

Required
cooling
capacity
(Btu/hr) _ Total cooling load

123,720 Btu/24 hr

Desired running time
_ 136,100 Btu/24 hr

16 hr
= 8,500 Btu/hr

Example 10-12. The dimensions of a
banana storage room located in New Orleans,
Louisiana are 22 ft x 32 ft x 9 ft. The walls
are 1 in. boards on both sides of 2 x 4 studs and
insulated with 3$ in. granulated cork. The floor
and roof are of the same construction as the
walls. The floor is over a ventilated crawl space
and the roof is exposed to the sun. The tempera-
ture outside the storage room is approximately
the same as the outdoor design temperature for
the region. 30,000 lb of bananas are ripened at
70° F and then cooled to 56° F in 12 hr for
holding storage. Compute the Btu/hr cooling
load. (Note: Because of the high storage tem-
perature the evaporator will not collect frost and
the equipment is designed for continuous run.)

COOLING LOAD CALCULATIONS 157

= 704 sq ft
= 1676sqft

- 5208 cu ft

=*89°F

= 0.079Btu/hr/sqft/°F

= 20°F

= 7/24hr

Solution
Outside surface area
Ceiling
Walls and floor

Inside volume
(21 ft x 31 ft x 8 ft)

Outside design
temperature for
New Orleans (Table
10-6)

Wall U factor
(Table 10-2)

Sun factor for roof
(Table 10-7) (tar
roofing)

Air changes (Table
10-9 A) (by interpo-
lation)

Heat gain per
cubic foot of air

change(Table 10-8A) = 1 .4 Btu/cu ft
(interpolated)

Specific heat of
bananas (Table
10-10)

Reaction heat of
bananas (Table
10-14)
Wall gain load:

Area x U x TD x 24 hr
Ceiling
704 x 0.079 x 53 x 24

= 70,740 Btu/24 hr
Walls and floor

1676 x 0.79 x 33 x 24

= 104,860 Btu/24 hr
Air change load:
Inside volume

x air changes x Btu/cu ft
= 5208 x 7 x 1.4 Btu/cu ft

= 51,000 Btu/24 hr
Product load:
Temperature reduction
_ M x C xjTj-Tj) x24hr

Chilling time (hr)
_ 30,000 x 0.9 x 14 x 24
12

= 756,000 Btu/24 hr
Respiration heat
= M x reaction heat x 24 hr
= 30,000 x 0.5 x 24 = 360,000 Btu/24 hr
Summation:

= 0.9Btu/lb/°F
= 0.5 Btu/lb/hr

Safety factor (10%)
Total cooling load

1,342,600 Btu/24 hr

1 34,260 Btu
1,476,860 Btu/24 hr

Required cooling capacity
(Btu/hr)

Total cooling load
Desired running time

_ 1,476,860 Btu/24 hr
24 hr

= 61,530 Btu/hr

Example 10-13. A chilling room 35 ft x
50 ft x 15 ft is used to chill 50,000 lb of fresh
beef per day from an initial temperature of
100° F to a final temperature of 35° F in 18 hr.
Four people work in the chiller during the
loading period. The lighting load is 1500 watts.
The floor, located over an unconditioned space,
is a 5 in. concrete slab insulated with 4 in. of
corkboard and finished with 3 in. of concrete.
The ceiling, situated beneath an unconditioned
space, is a 4 in. concrete slab with wood sleepers
and insulated with 4 in. of corkboard. All of
the walls are inside partitions adjacent to uncon-
ditioned spaces (90° F) except the east wall
which is adjacent to a 35° F storage cooler. Wall
construction is 4 in. cinder block insulated with
4 in. of corkboard and finished pn one side with
plaster. Compute the cooling load in Btu per
hour based on a 16-hr operating time for the
equipment.

Solution

Outside surface area
Ceiling
(35 ft x50ft)
Floor

(35 ft x50ft)
Walls (except
east)

(120 ft x 15 ft)
Inside volume
(34 ft x 49 ft x
13.5 ft)

Air changes
(Table 10-9A) (by
interpolation)

Heat gain per
cubic foot (Table
10-8A)(50%RH)

Specific heat of
beef (Table 10-12)
Chilling rate factor
(Table 10-12)
Occupancy heat
gain (Table 10-16)
Ceiling U factor
(Table 10-3)

= 1,750 sq ft
= 1,750 sq ft

= 1,800 sq ft

- 22,491 cu ft

= 3.2 per 24 hr

= 2.17 Btu/cu ft
= 0.75 Btu/lb/° F
= 0.67

= 900 Btu/hr/person
= 0.069 Btu/hr/sqft/°F

158 PRINCIPLES OF REFRIGERATION

Floor U factor

(Table 10-3) = 0.066 Btu/hr/sq ft/°F

Wall U factor

(Table 10-1) = 0.065 Btu/hr/sq ft/°F

Wall gain load:
A x U xTD x 24 hr
= 5300 x 0.067 x55 x24hr

= 468,700 Btu/24 hr
Air change load:
Inside volume

x air changes x Btu/cu ft
= 22,491 x 3.2 x 2.17

= 156,000 Btu/24 hr
Product load:

M x C x (T % - T x ) x 24
Chilling time (hr) x chilling rate factor
50,000 x 0.75 x 65 x 24
18 x 0.67

= 4,850,700 Btu/24 hr
Miscellaneous:
Occupancy

= No. of people x factor x 24 hr
= 4 x900 x24

= 86,400 Btu/24 hr
Lighting load
= watts x 3.4 x 24 hr
1500 x 3.4 x 24 = 122,400 Btu/24 hr
Summation: 5,684,200 Btu/24 hr

Safety factor (10%)

= 568,420 Btu
6,252,620 Btu/24 hr
Required cooling capacity (Btu/hr)

_ Total cooling load
Desired operating time
6,252,620 Btu/24 hr
18 hr
= 390,800 Btu/hr

Note: (1) Since there is no temperature
differential across the east wall, there is no gain
or loss of heat through the wall and the wall
is ignored in the cooling load calculations
However, this wall should be insulated with at
least the minimum amount of insulation to
prevent excessive heat gains through this wall
in the event that the adjacent refrigerated space
should become inoperative. (2) Since the TD
across all the walls, including floor and ceiling,
is the same and since the difference in the wall
U factors is slight, the walls may be lumped
together for calculation. (3) Although the
workmen are in the space for only 4 hr each
day for the purpose of load calculation, they are
assumed to be in the cooler continuously. This
is because their occupancy occurs simultan-

eously with the chilling peak. If the occupancy
occurred at any time other than at the peak,
the occupancy load could be ignored.

Example 10-14. Three thousand lug boxes
of apples are stored at 35° F in a storage cooler
50 ft x 40 ft x 10 ft. The apples enter the cooler
at a temperature of 90° F and at the rate of 200
lugs per day each day for the 15 day harvesting
period. The walls including floor and ceiling
are constructed of 1 in. boards on both sides of
2x4 studs and are insulated with 3$ in. of
rock wool. All of the walls are shaded and the
ambient temperature is 85° F. The average
weight of apples per lug box is 59 lb. The lug
boxes have an average weight of 4.5 lb and a
specific heat value of 0.60 Btu/lb/° F. Determine
the average hourly cooling load based on 16 hr
operating time for the equipment.

Solution
Outside surface
area = 5800 sq ft

Inside volume
(49 ft x 39 ft x 9 ft) = 17,200 cu ft

Wall U factor
(Table 10-2) = 0.072 Btu/hr/sq ft/ F

Air changes
(Table 10-9A) (by
interpolation) = 3.7 per 24 hr

Heat gain per
cubic foot (Table
10-8 A) = 1.86 Btu/cu ft

Specific heat of
apples (Table 10-10) = 0.89 Btu/lb/° F

Respiration heat
of apples (Table
10-14) (by interpo-
lation) = 0.025 Btu/lb/hr

Wall gain load:
A x U x TD x 24 hr
= 5800 x 0.072 x 50 x 24

= 501,100 Btu/24 hr
Air change load:
Inside volume x air changes

x Btu/cu ft
= 17,000 x 3.7 x 1.86

= 11 7,000 Btu/24 hr
Product load:
Temperature reduction
_ M x C x (T t - T x )
Chilling rate factor
Apples
(200 x 59 lb) x 0.89 x 55

0.67

= 862,100 Btu/24 hr

COOLING LOAD CALCULATIONS 159

Fig. 103

Lug boxes
(200 x 4.5 lb) x 0.6 x 55
0.67

= 44,300 Btu/24 hr
Respiration

= M x reaction heat x 24 hr
= (3000 x 59 lb) x 0.025 x 24

= 106,200 Btu/24 hr

Summation

Safety factor (10%)
Total cooling load

Average hourly load =

1,630,700 Btu/24 hr
= 163,100 Btu
= 1,793,800 Btu/24 hr
_ Total cooling load

Running time
_ 1,793,800 Btu/24 hr

16 hr
= 112,100 Btu/24 hr

Note: Load calculation and equipment
selection is based on maximum loading which
occurs on the fifteenth day.

Example 10-15. Twenty-two thousand
pounds of dressed poultry are blast frozen on
hand trucks each day (24 hr) in a freezing tunnel
14 ft x 9 ft x 10 ft high (see Fig. 10-3). The
poultry is precooled to 45° F before entering the
freezer where it is frozen and its temperature
lowered to 0° F for storage. The lighting load
is 200 watts. The hand trucks carrying the
poultry total 1400 lb per day and have a specific
heat of 0.25 Btu/lb/° F. The partitions adjacent
to the equipment room and vestibule are con-
structed of 6 in. clay tile insulated with 8 in. of
corkboard. Partitions adjacent to storage
cooler are 4 in. clay tile with 2 in. corkboard
insulation. The roof is a 6 in. concrete slab
insulated with 8 in. of corkboard and covered
with tar, felt, and gravel. The floor is a 6 in.
concrete slab insulated with 8 in. of corkboard

and finished with 4 in. of concrete. The floor is
over a ventilated crawl space. Roof is exposed
to the sun. The equipment room is well venti-
lated so that the temperature inside is approxi-
mately the outdoor design temperature for the
region. The storage room is maintained at 0° F,
whereas the temperature in the freezer is — 10° F.
The location is Houston, Texas. Determine the
average hourly refrigeration load based on 20 hr
per day operating time for the equipment.

Solution

Outside surface area
Roof

(9 ft + 14 ft) =126sqft

Floor

(9 ft x 14 ft) =126sqft

N and E partitions
(23 ft x 10 ft) =230sqft
S and W partitions
(23 ft x 10 ft) =230sqft
Inside volume
(8ft x9ft x 13ft)
Summer outdoor
design temperature
U factors
Roof

(Table 10-3)
Floor
(Table 10-3)

= 936 cu ft

92° F

= 0.036 Btu/hr/sq ft/° F

= 0.035 Btu/hr/sq ft/° F
N and E partitions

(Table 10-2) = 0.035 Btu/hr/sq ft/° F
S and W partitions

(Table 10-2) =0.12 Btu/hr/sq ft/° F

Roof sun factor
(Table 10-7)

Air changes
(Table 10-9B)

= 20°F

13.5 per 24 hr

Heat gain per cubic foot
(Table 10-8B) = 3.56 Btu/cu ft

160 PRINCIPLES OF REFRIGERATION

Cold
18' storage
38*F

— »-N
-32.5*-

Lochtrs

O'F

Lockers

Freezer

Lockers

N-

Cold storage
38"F

Conditioned
space, 80* F

Fig. 10-4. Frozen food locker plant.

Specific heat of poultry
(Table 10-12)

Above freezing

= 0.79 Btu/lb/° F
Below freezing

= 0.37 Btu/lb/° F
Latent heat of poultry
(Table 10-12)

= 106 Btu/lb
Freezing temperature

27° F
Wall gain load:
^xt7xri)x24hrs
Floor

126 x 0.035 x 102 x 24

= 10,800 Btu/24 hr
Roof
126 x 0.036 x (102 + 20) x 24

= 13,300 Btu/24 hr
South and west partition
230 x 0.035 x 102 x 24

= 19,700 Btu/24 hr
North and east partition
230 x 0.12 x 10 x 24

= 6,600 Btu/24 hr
Air change load:
Inside volume x air changes
x Btu/cu ft
- 936 x 13.5 x 3.56

= 45,000 Btu/24 hr
Product Load:
Temperature reduction
= M x C x(r s -TJ
Poultry

_ 22,000 x 0.79 x (45 - 27)
0.67 (chilling factor)

= 302,700 Btu/24 hr
22,000 x 0.37 x (27 - 0)

- 219,700 Btu/24 hr
Trucks

1,400 x 0.25 x (92 - 0)

Freezing = M x latent heat
Poultry = 22,000 x 106

= 2,332,000 Btu/24 hr
Miscellaneous:
Lighting: 200 watts x 3.4
Btu/watt/hr x 24 hr

= 16,300 Btu/24 hr

Summation:
Safety factor (10%)

Total cooling

load
Average hourly

load

3,014,100 Btu/24 hr
= 301,400 Btu

= 3,315,500 Btu/24 hr

_ 3,315,500 Btu/24 hr
20 hr (running time)
= 165,775 Btu/hr

Example 10-16. A frozen food locker plant
18 ft x 32.5 ft x 10 ft, containing 353 indi-
vidual lockers and an 8 ft freezing cabinet, is
located in Tulsa, Oklahoma (see Fig. 10-4).
The north and west wall are constructed of 8 in.
clay tile with 6 in. of corkboard insulation.
South and east walls are 4 in. clay tile with 4 in.
of corkboard insulation. The roof is exposed to
the sun and is constructed of 4 in. of concrete
insulated with 8 in. of corkboard and covered
with tar, felt, and gravel. The floor is a 5 in.
concrete slab poured directly on the ground,
insulated with 8 in. of corkboard, and finished
with 3 in. of concrete. The product load on
cabinet freezer is 700 lb of assorted meats per
day. (Standard practice is to allow for 2 lb of
product per locker per day.) In this instance,
the product is precooled to 38° F before being
placed in the freezer. The lighting load is 500
watts and the average occupancy is three people.
Determine the average hourly refrigerating rate
based on a 20-hr equipment operating time.

0.67

. 48,000 Btu/24 hr

Solution

Outside surface area
Roof (18

Floor (18

South and east
partition (50

North partition (18

West wall (32

Inside volume
(16 ft x 30.5 ft x 8

ft x 32.5 ft)
585 sq ft
ft x 32.5 ft)
585 sq ft

.5 ft x 10 ft)

505 sq ft

ft x 10 ft)

180 sq ft
.5 ft x 10 ft)

325 sq ft

ft)

3904 cu ft

COOLING LOAD CALCULATIONS 161

Design outdoor temperature
(Table 10-6) = 92°F

Design ground temperature
(Table 10-6A) = 65° F

Roof sun factor
(Table 10-7) = 20°F

West wall sun factor
(Table 10-7) - 6° F

U factors
Roof (Table 10-3)

= 0.036 Btu/hr/sq ft/ F

Floor (Table 10-3)

= 0.046 Btu/hr/sq ft/° F

North and west walls

(Table 10-1) = 0.034 Btu/hr/sq ft/° F
South and east walls

(Table 10-1) = 0.066 Btu/hr/sq ft/° F

Air changes
(see Note 2 of Table 10-9B)

= 12.6(6.3 x 2)

per 24 hr
Heat gain per cubic foot
(Table 10-8B) = 3.0 Btu/cu ft

Specific heat (average for meat)
Above freezing = 0.8 Btu/lb/° F
Below freezing = 0.4 Btu/lb/° F
Latent heat

(average) = 100 Btu/lb

Freezing tempera-
ture (average) = 28° F
Occupancy factor

= 1300 Btu/hr/person

Wall Gain Load:
A x U x TD x 24
Roof 585 x 0.036

x (92 + 20) x 24
= 56,600 Btu/24 hr
Floor 585 x 0.046 x 65 x 24

= 42,000 Btu/24 hr
South and east
partition 505 x 0.066 x 38 x 24

= 30,400 Btu/24 hr
North partition 180 x 0.034 x 80 x 24
= 11,750 Btu/24 hr

West wall 325 x 0.034

x (92 + 6) x 24
= 26,000 Btu/24 hr

Air change load:
Inside volume x air changes
x Btu/cu ft
= 3904 x 12.6 x 3.01

= 147,570 Btu/24 hr

= 531,560 Btu/24 hr

Product load:
Temperature reduction
= M x C x (T t - TJ
700 x 0.8 x (38 - 28)

= 5600 Btu/24 hr
700 x 0.4 x (28 - 0)

7840 Btu/24 hr
Freezing

= M x latent heat
= 700 x 100 70,000 Btu/24 hr

Miscellaneous:
Lights
= 500 watts x 3.4 Btu/hr

x 24 hr = 40,800 Btu/24 hr

Occupancy
= 3 x 1300 x 24

= 93,600 Btu/24 hr

Summation:
Safety factor

(10%) = 53,150 Btu

Total cooling

load = 584,710 Btu/24 hr

Average hourly
load _ 584,710 Btu/24 hr

20 hr
= 29,240 Btu/hr
Load on freezer 91,780 Btu/24 hr

20 hr
(product load
only, including
10% safety factor)

= 4,590 Btu/hr
Load on locker
room only = 29,240 Btu/hr

(total load less - 4,590 Btu/hr

freezer load) = 24,650 Btu/hr

Example 10-17. Five hundred gallons of
partially frozen ice cream at 25° F are entering
a hardening room 10 ft x 15 ft x 9 ft each day.
Hardening is completed and the temperature of
the ice cream is lowered to —20° F in 10 hr.
The walls, including floor and ceiling, are
insulated with 8 in. of corkboard and the over-
all thickness of the walls is 12 in. The ambient
temperature is 90° F and the lighting load is 300
watts. Assume the average weight of ice cream
is 5 lb per gallon, the average specific heat below
freezing is 0.5 Btu/lb/°F, and the average
latent heat per pound is 100 Btu.* Determine
the average hourly load based on 18 hr opera-
tion.

* These values are variable and depend upon the
composition of the mix, the percent of overrun,
and the temperature of the ice cream leaving the
freezer.

162 PRINCIPLES OF REFRIGERATION

Solution
Outside surface

area = 750 cu ft

Inside volume

(8 ft x 13 ft x 7 ft)

= 728 cu ft
99 Btu/sq ft/24 hr

Wall gain factor

(Table 10-18)
Air changes

(Table 10-9B)

(interpolated) = 16.7 per 24 hr

Heat gain per cubic foot
(Table 10-8B)
(50 % RH) =3.88 Btu/cu ft

Wall gain load:
area x wall gain factor
= 750 x 99 = 74,250 Btu/24 hr

Air change load:
Inside volume x air changes
x Btu/cu ft
= 728 x 16.7 x 3.88

= 47,170 Btu/24 hr
Product load:
Temperature reduction

M x C(T S - 7\) x 24 hr
— Chilling time

(500 x 5) x 0.5 x (25 20) x 24

10
= 135,000 Btu/24 hr
Freezing
_ M x latent heat x 24

Freezing time
_ (500 x 5) x 100 x 24
10

= 600,000 Btu/24 hr
Miscellaneous load:
Lighting:
300 watts x 3.4 x 24

= 24,480 Btu/24 hr

Summation: 880,900 Btu/24 hr

Safety factor (10%) = 88,090 Btu

968,990 Btu/24 hr
Average hourly load
(968,990/1 8 hr) = 53,800 Btu/hr

Example 10-18. A cooler 10 ft x 12 ft x 9
ft is used for general purpose storage in a
grocery store. The cooler is maintained at 35° F
and the service load is normal. The walls are
insulated with the equivalent of 4 in. of cork-
board and the ambient temperature is 80° F.
Determine the cooling load in Btu per hour
based on a 16 hr operating time.

Solution
Outside surface area

Inside volume
(9 ft x lift x 8 ft)

= 636 sq ft

Wall gain factor

(Table 10-18)
Usage factor

(Table 10-17)

Wall gain load:
area x wall gain factor
= 636 x 81

Usage load:
Inside volume x
usage factor
= 792 x 50

792 cu ft

81 Btu/sq ft/24 hr

50 Btu/cu ft/24 hr

51,500 Btu/24 hr

Total cooling load
Average hourly load

/ 91,100 Btu/2 4 hr

I 16 hr

= 39,600 Btu/24 hr
= 91,100 Btu/24 hr

5,700 Btu/hr

PROBLEMS

1. A cooler wall 10 ft by 18 ft is insulated with
the equivalent of 3 in. of corkboard. Compute
the heat gain through the wall in Btu/24 hr if the
inside temperature is 37° F and the outside
temperature is 78° F. Ans. 165,312 Btu/24 hr

2. The north wall of a locker plant is 12 ft by
60 ft and is constructed of 8 in. hollow clay tile
insulated with 8 in. of corkboard. The locker
plant is located in Houston, Texas and the
inside temperature is maintained at 0° F. Deter-
mine the heat gain through the wall in Btu/24 hr.

Ans. 54,000 Btu/24 hr

3. A cold storage warehouse in Orlando,
Florida has a 30 ft by 50 ft flat roof constructed
of 4 in. of concrete covered with tar and gravel
and insulated with the equivalent of 4 in. of
corkboard. If the roof is unshaded and the
inside of the warehouse is maintained at 35° F,
compute the heat gain through the roof in
Btu/24 hr. Ans. 181,330 Btu/24 hr

4. A small walk-in cooler has an interior volume
of 400 cu ft and receives heavy usage. If the
inside of the cooler is maintained at 35° F and
the outside design conditions are 90° F and 60%
relative humidity, determine the air change load
in Btu/24 hr. Ans. 50,300 Btu/24 hr

5. A frozen storage room has an interior volume
of 2000 cu ft and is maintained at a temperature

COOLING LOAD CALCULATIONS 163

of -10° F. The usage is light and the outside
design conditions (anteroom) are 50° F and 70 %
relative humidity. Compute the air change load
in Btu/24 hr. Arts. 16,100 Btu/24 hr

6. Five thousand pounds of fresh, lean beef
enter a chilling cooler at 100° F and are chilled
to 38° F in 24 hr. Compute the chilling load in
Btu/24 hr. Ans. 347,000 Btu/24 hr

7. Five hundred pounds of prepared, packaged
beef enter a freezer at a temperature of 36° F.

The beef is to be frozen and its temperature
reduced to 0° F in 5 hr. Compute the product
load. Ans. 267,900 Btu/hr

8. Fifty-five hundred crates of apples are in
storage at 37° F. An additional 500 crates enter
the storage cooler at a temperature of 85° F and
are chilled to the storage temperature in 24 hr.
The average weight of apples per crate is 60 lb.
The crate weighs 10 lb and has a specific value
of 0.6Btu/lb/°F. Determine the total product
load in Btu/24 hr. Ans. l,679,80aBtu/24 hr

Evaporators

1 1 — 1 . Types of Evaporators. As stated pre-
viously, any heat transfer surface in which a
refrigerant is vaporized for the purpose of
removing heat from the refrigerated space or
material is called an evaporator. Because of
the many different requirements of the various
applications, evaporators are manufactured in
a wide variety of types, shapes, sizes, and
designs, and they may be classified in a number
of different ways, such as type of construction,
operating condition, method of air (or liquid)
circulation, type of refrigerant control, and
application.

1 1-2. Flooded and Dry-Expansion Evapora-
tors. Evaporators fall into two general cate-
gories, flooded and dry expansion, according to
their operating condition. The flooded type is
always completely filled with liquid refrigerant,
the liquid level being maintained with a float

valve or some other liquid level control (Fig.
11-1). The vapor accumulating from the boiling
action of the refrigerant is drawn off the top
by the action of the compressor. The principal
advantage of the flooded evaporator is that the
inside surface of the evaporator is always
completely wetted with liquid, a condition that
produces a very high rate of heat transfer. The
principal disadvantage of the flooded evapo-
rator is that it is usually bulky and requires a
relatively large refrigerant charge.

Liquid refrigerant is fed into the dry-expan-
sion evaporator by an expansion device which
meters the liquid into the evaporator at a rate
such that all the liquid is vaporized by the time
it reaches the end of the evaporator coil (Fig.
11-2). For either type, the rate at which the
liquid is fed into the evaporator depends upon
the rate of vaporization and increases or
decreases as the heat load on the evaporator
increases or decreases. However, whereas the
flooded type is always completely filled with
liquid, the amount of liquid present in the dry-
expansion evaporator will vary with the load on
the evaporator. When the load on the evapo-
rator is light, the amount of liquid in the evapo-
rator is small. As the load on the evaporator
increases, the amount of liquid in the evaporator
increases to accommodate the greater load.
Thus, for the dry-expansion evaporator, the
amount of liquid-wetted surface and, therefore,
the evaporator efficiency, is greatest when the
load is greatest.

11-3. Types of Construction. The three
principal types of evaporator construction are:
(1) bare-tube, (2) plate-surface, and (3) finned.

Liquid from
receiver

Float control

Fig. Il-I. Flooded evaporator.
Notice accumulator and float
control. Circulation of the
refrigerant through the coil
is by gravity. The vapor accu-
mulated from the boiling action
in the coil -escapes to the top
of the accumulator and is
drawn off by the suction of the
compressor.

164

EVAPORATORS l*S

Bore-lube and plate-surface evaporators are
sometimes dawned together as prime-surface
evaporaion in thai the entire surface of both
these types u more or leu in contact with the
vaporiring refrigerant inside. With the finned
evaporator, the refrigcran(<cafTying tube are
the only prime surface. The fins themselves are
not rilled wilh Tcfrigerani and are, therefore,
only secondary hat transfer surfaces whose
function J* (o pick up heat from the surrounding
air and conduct it to the refrigerant-carrying
tubes.

Atthough prune-surface evaporate** of both
the bare-tube and plait-surface types give salt*
factory service on a wide variety of applications
operating in any temperature range, they are
most frequently applied to applications where
the space temperature is maintained below
34* F and frost accumulation on the evaporator
surface cannot be readily prevented. Frost
accumulation on prime-surface evaporaion
does not affect the evaporator capacity to use
extent thai it doc* on finned coils. Further'
more, most prime surface evaporators, particu-
larly the fiate-turface type, are easily cleaned
and can be readily defrosted manually by either
brushing or scraping off the frost accumulation.
This can be accomplished without interrupting
(be refrigerating process and jeopardizing the
quality of the refrigerated product.
(ML ftar*~Tub* Evaporator*. Bare- lube
evaporators a re usually constructed of cither steel
pipe or copper tubing. Steel pipe is used for
luge evaporators and for evaporators to be
employed with ammonia, whereas copper tubing
it utilized in the manufacture of smaller evepo*

Liquid *nm Rtfnetrtnt

ntctiw j^Ro* contra*

Mr Mb

Fig. 11-5. Dry-wpvtslun aviporuer. Liquid r#-
frlftrtftl refwrim prof r«tii inly u It flown through
OOll Ifld l»*yet toil u i vapor. F«cler bulb cpnireU
rata of now through tha orrflca at tha flow control.

< U 'c

(a)

d6a

Flf. 11-3. Common dmgm far b*rt-tub* «H».
(o> fin cignf eo+l. {&} C»*»l trwnewi* coil.

niton intended for use with refrigerants other
than ammonia. Bare-tube coils are available
tn a number of sues, shapes, and designs, and
are usually custom made lo the individual
application. Common shapes for bare-tube
coils are fiat zigzag and oval trombone, as
shown in Fig. I!-), Spiral bare-tube coils are
often employed for liquid chilling.
11-5. Plate-Surface Evaporators. Plale-sur*
face evaporators arc of several lypes. Some are
constructed of two fiat sheets of metal so
embossed and welded together as to provide a
path for refrigerant flow between the two sheets
(Fig. 11-4). This particular type of plate-
■urfacc evaporator is widely uaed in household
refrigerators and home freezers because it is
easily cleaned, economical to manufacture, and
can be readily formed into any one of the various
shapes required (Fig, 11-5).

(M PRINCIPLES OF REFRIGERATION

Flf. MM. Standard lerpenllne plate evaporator, (Courtety Kold-Hold Division — Truuer Manufacturing, Inc.)

■ ^^ r mmm

Rg. 11-5. Soma typical iheocs «raJl*b!e in pfaie-typ* eviporiion. (Courtesy D«s- Product*. Inc.)

(A) Outside jacket Of plate. Heavy, olnctrkally (E) Fitting whir* vacuum ll drawn and than permajv
wddfid steel. Smooth surface, ontlv (tiled.

(B) Continuous steel tubing through which refrlg- (F) Vacuum ip«« In dry pit to Spice in hold-over
eratit passei. p.| K( contains lutectk solution under vacuum. No
{C) Inlet from compressor, maintenance required due. CO i!urdy, limp In con-
(D> Outfit to compressor. Copper connections for struct Jon. No moving parti; no th ing to wear or get
all refrigerant* except ammonia where stiel cormee- out of order; no service neeusiry.

lions ere vied.

Fig. II-*, Plate-type evapc-retor, (Courtesy Dolt Refrigerating Company.}

EVAPORATORS

1*7

Another type of plalc-surfacc evaporator
consuls of formed tubing installed between two
metal plates which arc welded together at the
edges (Fig. It -6). In order to provide good
thermal contact between the welded plated tnd
the tubing carrying the refrigerant, live space
between the plates is either filled with a eutcciic
solution or evacuated so that the pressure of the
atmosphere exerted on the outside surface of
the plates holds the plates firmly against the
tubing inside. Those containing the eutectic
Solution are especially useful where a holdover
capacity is required. Many are used on refriger-
ated trucks, let such applications, the plates are
mounted either vertically or horizontally from
the ceiling or walls of the truck { Fig. 1 1-7) and
are usually connected to a central plant refriger-
ation system while the trucks are parked at the
terminal during the night. The refrigerating
capacity thui stored in the eutectic solution is
sufficient to refrigerate the product during the
oral day's operations. The temperature of the

plates is controlled by the melting point of the
eutectic solution.

Plate-type evaporators may be used singly or
in banks. Figure 11-8 illustrates how the plates
can be grouped together for ceiling mounting in
holding rooms, locker plants, freezers, etc. The
plates may be manifolded for parallel flow of
the refrigerant (Fig. 11-9) or they may be
connected for series flow,

Plate-surface evaporators provide excellent
shelves in freezer rooms and similar applications
(Fig. 11-10). They are also widely used a*
partitions in freezers, frozen food display cases,
Ice cream cabinets, uhIji fountains, etc. Plate
evaporators are especially useful for liquid
cooling installations where unusual peak load
conditions are encountered periodically. By
building up an ice bank on the surface of the
plain during periods of light loads, a holdover
refrigerating capacity is established which will
help i be refrigerating equipment carry the load
through the heavy or peak conditions Fig, 11-11),

Pig, 1 1*1. Frtmitr plitu Iniullad In whottHlt Fc* crttm truck body, (Court aty Do4« H«tri| •r»tln| Company.)

IfiS PRINCIPLES OF REFRIGERATION

Flj. 11-$. P1»te b»nks employed in I o w MfnpcnLu r* itor-ifi roomt. (Ccmttmy Dqfa ftcfrt|enuifl| Company.)

Since this allows Che use of smaller capacity
equipment than would ordinarily be required
by the peak load, a savings « affected in Initial
cost and, usually, also in operating expenses,
1 1-6. Finned Evaporators. Finned eoib in
bare- tube coils upon which metal plates or fins
have been insialled (Fig, 13-15), The fins,
serving as secondary heat-absorbing surfaces,
have the effect of increasing the over-all surface
area of the evaporator, thereby improving its
efficiency. With bare-lube evaporators, much
of the air that circa kites over the coil passes
through the open spaces between Ihe tubes and
docs not come in contact with the cm I surface.
When fins arc added to a coil, the fins extend
out imo the open spaces be (ween the tubes and
act as heat collectors. They remove heat from
that portion of the air which would not ordin-
arily conic in contact with the prime surface
and conduct it back to the tubing,

It is evident that to be effective the fins must

be connected to the tubing in such a manner
that good thermal contact between the fins and
the tubing is assured. In some instances, the
flni are soldered directly to [he i ubing. In others,
the fins are slipped over ihe tubing and the
tubing is expanded by pressure or some such
means so that the fins bite imo the tube surface
and establish good thermal contact. A variation
of the latter method is to flare the fin hole
slightly to allow the fin to slip over the tube.
After the fin is installed, the flare is straightened
and the fin is securely locked to the tube.

Fin Ni« and spacing dej^eiid in part on the
particular type of application for which the coil
is designed, The size of the tube determines the
size of the fin. Small tubes require small fins.
As the size of the tube increases, the size of the
fin may be effectively increased, Fin spacing
varies from one to fourteen fins per inch,
depending primarily on the operating tempera-
ture of the coil.

fVAfORATOAS

16?

Fl|» 1 1-*. Ptau bank with pluu minifeJdtd for ptraJM rrfriftnr* ftow. Plat** may ttto b« Kmn«t*d for
ien« tlow. (Cduntiy Kold-Hdd Dickon— Trinwr H»nuf»ctur1ni, '"t-J

Fig. tl-10. P1*» fnpQjiwn
implayid u frtenr indvti.
Not* thu plitM »r* imngtdi
for Mhti ndrtftnni flow.
{Courtesy KoJd-Hold Division
— Trwwr Minuhcturinf , Jnc)

170 PRINCIPLES OF AEFAKJEAATTON

Ft t . 11.11. let-Cat Rtfrttar*-
tion hoMov«r- capacity It etnb-
li*h*d by buii^.ftf up ■ tank
of l« on plite «aipenun,
(Court**)* Dol* Rafrlfarilinf

Cwnptflf.)

Frost accumulation on air-cooling colls
operating at low temperatures is unavoidable,
and since any frost accumulation on finned coils
tends to restrict the air passages between the
lint and to retard air circulation through the
coil, evaporators designed for tow temperature
application! must have wide fin spacing (two
or Ihree fins per inch) in order to minimize the
danger of blocking air circulation. On the
other hand, coils designed for air conditioning
and other installation* where the coil Optra ICi
it temperatures high enough so that no frost
accumulates On the coil surface may have as
many as fourteen fins per inch.

When air circulation over finned coils is by
gravity, it is important that the coil offer as little
resistance lo air How as is possible; therefore,
in general, fin spacing should be wider for
natural convection coil* than for coils employing
fans.

It has been determined that a definite relation-
ship exists between the inside and outside
surfaces of an evaporator . Since external finning
affects only the outside surface, I he addition of
(ins beyond a certain limit will not materially
increase the capacity of the evaporator. In
fad. in some instances, excessive finning may
actually reduce the evaporator capacity by

restricting the air circuit son over the coil
unnecessarily.

Since their capacity is affected more by frost
accumulation than any other type of evaporator,
finned coils ore best suited to air-cooling applica-
tion where the temperature is maintained above
34' F. When tinned soils are used for low
temperature operation, some means of defrost*
ing the coil at regular inter val* must be provided.
This may be accomplished automatically by
several means which are discussed in another
chapter ,

Because of the fins, finned coils have more
surface area per unit of length and width than
prime-surface evaporators and can therefore
be built more compactly. Generally, a finned
coil will occupy less space than either a bare-
lube or plate-surface evaporator Of* the same
capacity. Hut provides for a considerable
savings in space and makes tinned coils ideally
suited for use with funs a* In reed convection
units.

1 1-7. Evaporator Capacity. The capacity of
any evaporator or cooling coil is the rate at
which heat will pass th rough the evaporator
walls from the refrigeruk-n ipiioe or product to
the vaporizing liquid inside and is usually
expressed in Blu per hour An evaporator

selected for any specific application must always
have sufficient capacity to allow the vaporizing
refrigerant to absorb heat at the rate necessary
to produce the required cooling when operating
at the design conditions.

Heat reaches the evaporator by all three
methods of heat transfer. In air-cooling applica-
tions most of the heat is carried to the evapo-
rator by convection currents set up in the
refrigerated space either by action of a fan or by
gravity circulation resulting from the difference
in temperature between the evaporator and the
space. Too, some heat is radiated directly to
the evaporator from the product and from the
wall of the space. Where the product is in
thermal contact with the outer surface of the
evaporator, heat is transferred from the product
to the evaporator by direct conduction. This
is always true for liquid cooling applications
where the liquid being cooled is always in
contact with the evaporator surface. However,
circulation of the cooled fluid either by gravity
or by action of a pump is still necessary for
good heat transfer.

Regardless of how the heat reaches the out-
side surface of the evaporator, it must pass
through the walls of the evaporator to the
refrigerant inside by conduction. Therefore,
the capacity of the evaporator, that is, the rate
at which heat passes through the walls, is deter-
mined by the same factors that govern the rate
of heat flow by conduction through any heat
transfer surface and is expressed by the formula

Q =A x U x D (11-1)

where Q = the quantity of heat transferred in
Btu/hr
A = the outside surface area of the
evaporator (both prime and finned)
U = the over-all conductance factor in
Btu/hr/sq ft of outside surface/" F D
D = the logarithmic mean temperature
difference in degrees Fahrenheit
between the temperature outside the
evaporator and the temperature of
the refrigerant inside the evaporator
11-8. U or Over-All Conductance Factor.
The resistance to heat flow offered by the evapo-
rator walls is the sum of three factors whose
relationship is expressed by the following:

1 _R L 1

U~fi + K + f

(11-2)

EVAPORATORS 171

where V = the over-all conductance factor in

Btu/hr/sq ft/ F D
ft = the conductance factor of the inside

surface film in Btu/hr/sq ft of inside

surface/" F D
LjK = resistance to heat flow offered by

metal of tubes and fins
/„ =.the conductance factor of the out-
side surface film in Btu/hr/sq ft of

outside surface/" F D
R = ratio of outside surface to inside

surface

Since a high rate of heat transfer through the
evaporator walls is desirable, the U or conduct-
ance factor should be as high as possible.
Metals, because of their high conductance
factor, are always used in evaporator construc-
tion. However, a metal which will not react
with the refrigerant must be selected. Iron, steel,
brass, copper, and aluminum are the metals
most commonly used. Iron and steel are not
affected by any of the common refrigerants,
but are apt to rust if any moisture is present in
the system. Brass and copper can be used with
any refrigerant except ammonia, which dis-
solves copper. Aluminum may be used with
any refrigerant except methyl chloride. Mag-
nesium and magnesium alloys cannot be used
with the fluorinated hydrocarbons or with
methyl chloride.

Of the three factors involved in Equation
11-2, the metal of the evaporator walls is the
least significant. The amount of resistance to
heat flow offered by the metal is so small,
especially where copper and aluminum are
concerned, that it is usually of no consequence.
Thus, the U factor of the evaporator is deter-
mined primarily by the coefficients of con-
ductance of the inside and outside surface
films.

In general, because of the effect they have on
the inside and outside film coefficients, the value
of U for an evaporator depends on the type
of coil construction and the material used, the
amount of interior wetted surface, the velocity
of the refrigerant inside the coil, the amount
of oil present in the evaporator, the material
being cooled, the condition of the external
surface, the fluid (either gaseous or liquid)
velocity over the coil, and the ratio of inside
to outside surface.

172 PRINCIPLES OF REFRIGERATION

Heat transfer by conduction is greater through
liquids than through gases and the rate at which
the refrigerant absorbs heat from the evaporator
walls increases as the amount of interior wetted
surface increases. In this respect, flooded
evaporators, since they are always completely
filled with liquid, are more efficient than the
dry-expansion type. This principle also applies
to the external evaporator surface. When the
outside surface of the evaporator is in direct
contact with some liquid or solid medium, the
heat transfer by conduction to the outside
surface of the evaporator is greater than when
air is the medium in contact with the evaporator
surface.

Any fouling of either the external or internal
surfaces of the evaporator tends to act as ther-
mal insulation and decreases the conductance
factor of the evaporator walls and reduces the
rate of heat transfer. Fouling of the external
surface of air-cooling evaporators is usually
caused by an accumulation of dust and lint
from the air which adheres to the wet coil
surfaces or by frost accumulation on the coil
surface. In liquid-cooling applications, fouling
of the external tube surface usually results from
scale formation and corrosion. Fouling of the
internal surface of the evaporator tubes is
usually caused by excessive amounts of oil in
the evaporator and/or low refrigerant velocities.
At low velocities, vapor bubbles, formed by the
boiling action of the refrigerant, tend to cling
to the tube walls, thereby decreasing the amount
of interior wetted surface. Increasing the re-
frigerant velocity produces a scrubbing action on
the walls of the tube which carries away the oil
and bubbles and improves the rate of heat flow.
Thus, for a given tube size, the inside film coeffi-
cient increases as the refrigerant velocity in
creases. The refrigerant velocity is limited, how-
ever, by the maximum allowable pressure drop
through the coil and, if increased beyond a certain
point, will result in a decrease rather than an in-
crease in coil capacity. This depends to some
extent on the method of coil circuiting and is dis-
cussed later. It can be shown also that the con-
ductance of the outside surface film is improved
by increasing the fluid velocity over the outside
surface of the coil. But, here again, in many
cases the maximum velocity is limited, this time
by consideration other than the capacity of the
evaporator itself.

Any increase in the turbulence of flow either
inside or outside the evaporator will materially
increase the rate of heat transfer through the
evaporator walls. In general, internal turbu-
lence increases with the difference in temperature
across the walls of the tube, closer spacing of
the tubes, and the roughness of the internal tube
surface. In some instances, heat transfer is
improved by internal finning.

Outside flow turbulence is influenced by fluid
velocity over the coil, tube spacing, and the
shape of the fins.

11-9. The Advantage of Fins. The advantage
of finning depends on the relative values of the
coefficients of conductance of the inside and
outside surface films and upon R, the ratio of
the outside surface to the inside surface. In any
instance where the rate of heat flow from the
inside surface of the evaporator to the liquid
refrigerant is such that it exceeds the rate at
which heat passes to the outside surface from
the cooled medium, the over-all capacity of the
evaporator is limited by the capacity of the
outside surface. In such cases, the over-all
value of U can be increased by using fins to
increase the outside surface area to a point such
that the amount of heat absorbed by the outside
surface is approximately equal to that which
can pass from the inside surface to the liquid
refrigerant.

Because heat transfer is greater to liquids than
to vapors, this situation often exists in air-
cooling applications where the rate of heat flow
from the inside surface to the liquid refrigerant
is much higher than that from the air to the
outside surface. For this reason, the use of
finned evaporators for air-cooling applications
is becoming more and more prevalent. On the
other hand, in liquid-cooling applications, since
liquid is in contact with both sides of the evapo-
rator and the rate of flow is approximately equal
for both surfaces, barenube evaporators per-
form at high efficiency and finning is usually
unnecessary. In some applications, where fluid
velocity over the outside of the evaporator is
exceptionally high, the flow of heat to the outer
surface may be greater than the flow from the
inner surface to the refrigerant. When this
occurs, the use of inner fins will improve evapo-
rator capacity in that the amount of interior
wetted surface is increased. Several methods
of inner finning are shown in Fig. 11-12.

EVAPORATORS

173

(a)

(b)

Fig. 11-12. Some methods of
inner finning.

11-10. Logarithmic Mean Temperature
Difference. As illustrated in Fig. 11-13, the
temperature of air (or any other fluid) decreases
progressively as it passes through the cooling
coil. The drop in temperature takes place along
a curved line (A) approximately as indicated.
Assuming that the temperature of the refrigerant
remains constant, it is evident that the difference
in temperature between the refrigerant and the
air will be greater at the point where the air
enters the coil than at the point where it leaves,
and that the average or mean difference in
temperature will fall along the curve (a) at a
point somewhere between the two extremes.
Although the value obtained deviates slightly
from the actual logarithmic mean, an approxi-
mate mean temperature difference may be
calculated by the following equation:

D JU-t T )^H-Q (U3)

where D = the arithmetic mean temperature
/„ = the temperature of the air entering

the coil
fj = the temperature of the air leaving

the coil
t r = the temperature of the refrigerant

in the tubes

For the values given in Fig. 11-13, the arith-
metic mean temperature difference is

(40 - 20) + (30 - 20)

= 15° F

It must be remembered that the MTD as
calculated by Equation 11-3 is slightly in error
because of the curvature of the curved line A
and would be the actual MTD only if the drop
in air temperature occurred along a straight line,
as indicated by the dotted line B in Fig. 11-13.

The actual logarithmic mean temperature,
which is the midpoint of the curved line A, is
given by equation

(f. - t r ) - (/ x - t r )

D =

it. - t r )

0l - t r )

(11-4)

For the values given in Fig. 1 1 - 1 3, the logarith-
mic mean temperature difference will be

D =

(40 - 20) - (30 - 20)

40-20
30 -20

JO
ln2

= — = 14.43° F

The preceding calculations were made on the
assumption that the refrigerant temperature
remains constant. When this is not die case,

174 PRINCIPLES OF REFRIGERATION

Leaving air
temperature-30 F

Fig. 11-13. Mean temperature of air passing through
evaporator.

The velocity of the air passing over the coil
has a considerable influence on both the value
of U and the METD and is important in deter-
mining evaporator capacity. When air velocity
is low, the air passing over the coil stays in
contact with the coil surface longer and is
cooled through a greater range. Thus, the
METD and the rate of heat transfer is low. As
air velocity increases, a greater quantity of air
is brought in contact with the coil per unit of
time, the METD increases, and the rate of heat
transfer improves. In addition, high air
velocities tend to break up the thin film of
stagnant air which is adjacent to all surfaces.
Since this film of air acts as a heat barrier and
insulates the surface, its disturbance increases
the conductance of the outside surface film and
the over-all value of U improves.

t r will have two values. This condition is
discussed in another chapter.

The log mean temperature difference here-
after called mean effective temperature difference
(METD), may also be determined from Table
11-1.

11-11. The Effect of Air Quantity on
Evaporator Capacity. Although not a part of
the basic heat transfer equation, there are other
factors external to the coil itself which greatly
affect coil performance. Principal among these
are the circulation, velocity, and distribution of
air in the refrigerated space and over the coil.
These factors are closely related and in many
cases are dependent one on the other.

Except in liquid cooling and in applications
where the product is in direct contact with the
evaporator, most of the heat from the product
is carried to the evaporator by air circulation:
If air circulation is inadequate, heat is not
carried from the product to the evaporator at a
rate sufficient to allow the evaporator to per-
form at peak efficiency. It is important also
that the circulation of air is evenly distributed
in all parts of the refrigerated space and over
the coil. Poor distribution of the circulating
air results in uneven temperatures and "dead
spots" in the refrigerated space, whereas the
uneven distribution of air over the coil surface
causes some parts of the surface to function less
efficiently than others and lowers evaporator
capacity.

CoflA

CoilB

■Air

CoilC

Fig. 1 1-14. Coils B and C both have twice the surface
area of coil A. Coil C has twice the face area of coil A
or coil B.

EVAPORATORS 175

11-12. Surface Area. Equation 11-1 indicates
that the capacity of an evaporator varies
directly with the outside surface area. This
is true only if the U factor of the evaporator
and the METD remains the same. In many
cases, the value of U and the METD are affected
when the surface area of the evaporator is
changed. In such cases, the capacity of the
evaporator does not increase or decrease in
direct proportion to the change in surface area.
To illustrate, in Fig. 11-14, coils B and C each
have twice the surface area of coil A, yet the

ing the number of rows as in coil B, the METD
will be decreased and the increase in capacity
will not be nearly as great as when the surface
area is increased as in coil C. For the same
total surface area, a long, wide, flat coil will, in
general, perform more efficiently than a short,
narrow coil having more rows depth. However,
in many instances, the physical space available
is limited and compact coils arrangements must
be used. In applications where it it is permis-
sible, the loss of capacity resulting from increas-
ing the number of rows can be compensated for

3rd
Row

Fig. 11-15. Air temperature
drop across typical three-row
cooling coil.

Leaving -<-

Air -*r-

Temperature -<-

30*-«-

increase in capacity over the capacity of coil A
will be greater for coil C than for coil B. Pro-
vided the air velocity is the same (the total
quantity of air circulated over coil C must be
twice that circulated over coil A), the METD
across C will be exactly the same as that across
A and the capacity of C will therefore be twice
the capacity of coil A.

Figure 11-15 shows how the METD is af-
fected when the surface area of the coil is
increased by increasing the number of rows
(depth). Note that the drop in air temperature
is much greater across the first row and dimin-
ishes as the air passes across each succeeding
row. This is accounted for by the fact that the
temperature difference between the air and the
refrigerant is much greater across the first row,
becomes less and less as the temperature of the
air is reduced in passing across each row, and
is least across the last row. It is evident then
that the rate of heat transfer is greater for the
first row and that the first row performs the
most efficiently. For this reason, if the surface
area of coil A in Fig. 11-14 is doubled by increas-

2nd

Row

Air

32°

Air

C\J

1st

Row

■<

Air

35°

Air

Entering
Air

Temperature
40*

Coil

to some extent by increasing the air velocity over
the coil. Too, in some applications, the use of
deep coils is desirable for the purpose of
dehumidification.

11-13. Evaporator Circuiting. It was demon-
strated in Chapter 8 that excessive pressure drop
in the evaporator results in the suction vapor
arriving at the suction inlet of the compressor
at a lower pressure than is actually necessary,
thereby causing a loss of compressor capacity
and efficiency.

To avoid unnecessary losses in compressor
capacity and efficiency, it is desirable to design
the evaporator so that the refrigerant experi-
ences a minimum drop in pressure. On the
other hand, a certain amount of pressure drop
is required to flow the refrigerant through the
evaporator, and since velocity is a function of
pressure drop, the drop in pressure must be
sufficient to assure refrigerant velocities high
enough to sweep the tube surfaces free of vapor
bubbles and oil and to carry the oil back to the
compressor. Hence, good design requires that
the method of evaporator circuiting be such

176 PRINCIPLES OF REFRIGERATION

Refrigerant
out

Fig. 11-16. Evaporator with one series refrigerant
circuit.

that the drop in pressure through the evaporator
is the minimum necessary to produce refrigerant
velocities sufficient to provide a high rate of heat
transfer and good oil return.

In general, the drop in pressure through any
one evaporator circuit will depend upon the size
of the tube, the length of the circuit, and the
circuit load. By circuit load is meant the time-
rate of heat flow through the tube walls of the
circuit. The circuit load determines the quan-
tity of refrigerant which must pass through the
circuit per unit of time. The greater the circuit
load, tiie greater must be the quantity of
refrigerant flowing through the circuit and the
greater will be the drop in pressure. Hence,
for any given tube size, the greater the load on
the circuit, the shorter the circuit must be in
order to avoid excessive pressure drop.

Evaporators having only a single series
refrigerant circuit, such as the one illustrated
in Fig. 11-16, will perform satisfactorily within
certain load limits. When the load limit is
exceeded, the refrigerant velocity will be in-
creased beyond the desired range and the
pressure drop will be excessive.

Notice that the refrigerant enters at the top
of the evaporator as a liquid and leaves at the
bottom as a vapor. Since the volume of the
refrigerant increases as the refrigerant vaporizes,
the refrigerant velocity and the pressure drop
per foot increase progressively as the refrigerant
travels through die circuit, and are greatest at
the end of the coil where the refrigerant is 100 %
vapor.

The excessive pressure drop occurring in the
latter part of a single series circuit evaporator

can be eliminated to a great extent by splitting
the single circuit into two circuits in the lower
portion of the evaporator (Fig. 11-17). When
this is done, the refrigerant travels a single series
path until the refrigerant velocity builds to the
allowable maximum, at which time the circuit
is split into two parallel paths for the balance of
the travel through the evaporator. This has the
effect of reducing the refrigerant velocity in the
two paths to one-half the value it would have
without the split, and the pressure drop per foot
is reduced to one-eighth of the value it would
have in the lower part of the evaporator with a
single single series circuit.* This, of course,
will permit greater loading of the coil without
exceeding the allowable pressure drop. At the
same time, the velocity in all parts of the coil
is maintained within the desirable limits so
that the rate of heat transfer is not unduly
affected.

Another method of reducing the pressure drop
through the evaporator is to install refrigerant
headers at the top and bottom of the evaporator
so that the refrigerant is fed simultaneously
through a multiple of parallel circuits (Fig.
11-18). However, this arrangement is not too
satisfactory and is not widely used. While the
pressure drop through the evaporator is low,
this method of circuiting ordinarily results in

Fig. 11-17. Evaporator with split refrigerant circuit.

* Pressure drop increases as the square of the
velocity. Reducing the velocity to one-half reduces
the pressure drop to one-quarter of its original value.
Then, since the length of each parallel branch is
only one-half the length of a single circuit, the drop
in pressure in the lower portion of the split coil is
only one-eighth of the single circuit value.

EVAPORATORS 177

reducing the refrigerant velocity below the
desired minimum so that the inside film coefficient
and the rate of heat transfer are also low.
Another disadvantage of this type of circuiting
is that the loading of the circuits is uneven. Since
the temperature difference between the air
passing over the coil and the refrigerant in the
tubes is much greater across the first circuit (first
row) than across the last circuit (last row), the
loading of the first circuit is much greater than
the loading of the last circuit. Hence, the
refrigerant velocity and the drop in pressure
through the several circuits are uneven and a
large portion of the coil operates inefficiently.
This criticism can be applied to some extent
also for the circuit arrangements in Figs. H-16
and 1 1-17. In all three arrangements, the lower
portion of the evaporator will not perform as
effectively as the upper portion because wetting
of the internal tube surface will not be as great
in the lower portion. This is because the refrig-
erant in the lower portion contains a high
percentage of vapor, whereas in the upper
portion the refrigerant is nearly all liquid.

It is for this reason also that the outside
surface temperature of the coil is always lowest
near the refrigerant inlet and highest near the
outlet, in spite of the fact that the saturation
temperature of the refrigerant is lowest at the
outlet due to the drop in pressure through the
coil.

The circuit arrangement shown in Fig. 11-19
is very effective and is widely used, particularly
when circuit loading is heavy, as in the case of

Refrigerant

out V^

Refrigerant

Circuits

Fig. 11-18. Four-circuit evaporator with refrigerant
headers on both inlet and outlet. Crossflow of air
and refrigerant results in uneven circuit loading.

Refrigerant

Refrigerant
distributor

. > Sk. Refrigerant
out

Fig. 11-19. Evaporator with refrigerant distributor
and suction header. Notice counterflow arrangement
for refrigerant and air.

an air conditioning coil where the temperature
differential between the refrigerant and the air
is large and where external finning is heavy.
Notice that the air passes in counterflow to the
refrigerant so that the warmest air is in contact
with the warmest part of the coil surface. This
provides the greatest mean temperature differ-
ential and the highest rate of heat transfer.
Notice also that loading of the circuits is even.
The number and length of the circuits that such
a coil should have are determined by the size
of the tube and the load on the circuits.

For the multipass, headered evaporator, the
arrangement shown in Fig. 11-20 is much more
desirable than that shown in Fig. 11-18. Coun-
terflowing of the air and the refrigerant increases
the METD and permits more even loading of
the circuits.

1 1-14. Use of Manufacturer's Rating Tables.
The mathematical evaluation of all the factors
which influence evaporator capacity is usually
impractical and in many cases impossible. For
the most part, evaporator capacities must be
determined by actual testing of the evaporator.
The results of such tests are contained in the
rating tables published by the various evapo-
rator manufacturers.

The method of rating evaporators varies
somewhat with the type of evaporator and with
the particular manufacturer involved. How-
ever, the various rating methods are very similar
and most manufacturers include, along with the
evaporator rating tables, instructions as to how
to use the ratings. In most cases, where the

178 PRINCIPLES OF REFRIGERATION

Fig. 11-20. Counter-flow of
refrigerant and air results in
more even circuit loading and
a higher mean temperature
differential. Compare this
arrangement with the cross-
flow arrangement in Fig. 11-18.

evaporators are rated in accordance with ASRE
Standards, the capacity data are reliable and
are for operating conditions as normally en-
countered.

The selection of evaporators from manufac-
turer's rating tables is relatively simple once the
conditions at which the evaporator is to operate
are known. Typical evaporator rating tables,
along with methods of evaporator selection, are
discussed at the end of the chapter.
11-15. Evaporator TD. One of the most
important factors to consider in selecting the
proper evaporator for any given application
is the evaporator TD. Evaporator TD is
defined as the difference in temperature between
the temperature of the air entering the evapo-
rator and the saturation temperature of the
refrigerant corresponding to the pressure at the
evaporator outlet.*

Although more exact methods of rating
evaporators are necessary in order to select
evaporators for air conditioning applications
and some product storage applications where
space temperature and humidity are especially
critical, ratings for most evaporators designed
for product cooling applications are based on
evaporator TD.

The relationship between evaporator capacity
and evaporator TD is shown by the curve in
Fig. 11-21. Notice that the capacity of the
evaporator (Btu/hr) varies directly with the
evaporator TD. That is, if an evaporator has a
certain capacity at a 1 ° F TD, it will have exactly
ten times that capacity if the TD is increased to

* ASRE Standard 25-56, Methods of Rating Air
Coolers For Refrigeration.

10° F, provided that all other conditions are the
same.t

It is evident that a coil with a relatively small
surface area operating at a relatively large TD
can have the same capacity as another coil
having a larger surface area but operating at a
smaller TD. Thus, insofar as Btu per hour
capacity alone is concerned, a small coil will
have that same refrigerating effect as a larger
one, provided that the TD at which the small
coil operates is greater in proportion. However,
it will be shown in the following sections that
the temperature difference between the evapo-
rator and the refrigerated space has considerable
influence on the condition of the stored product
and upon the operating efficiency of the entire
system, and is usually, therefore, the determining
factor in coil selection. Before an evaporator
can be selected, it is necessary to first determine
the TD at which it is expected to function. Once
the desired temperature difference is known, an

t Care should be taken not to confuse METD
with evaporator TD. According to Equation 11-1,
the Btu per hour capacity of any given evaporator
(whose U factor and surface area are fixed at the
time of manufacture) varies directly with the
METD. However, assuming that the refrigerant
temperature and all else remains constant, the
METD between the air passing over the evaporator
and the refrigerant in the evaporator will vary
directly with the temperature of the air entering the
evaporator. That is, if the temperature of the air
entering the evaporator increases, the METD
increases. Hence, the METD varies in proportion
to the evaporator TD and, therefore, the capacity
of the evaporator also varies in proportion to the
evaporator TD.

evaporator having sufficient surface area to
provide the required cooling capacity at the
design TD can be selected.
11-16. The Effect of Coil TD on Space
Humidity. The preservation of food and other
products in optimum condition by refrigeration
depends not only upon the temperature of the
refrigerated space but also upon space humidity.
When the humidity in the space is too low,
excessive dehydration occurs in such products
as cut meats, vegetables, dairy products, flowers,
fruits, etc. On the other hand, when the humid-
ity in the refrigerated space is too high, the
growth of mold, fungus, and bacteria is encour-
aged and bad sliming conditions occur, particu-
larly on meats and especially in the wintertime.
Space humidity is of little importance, of course,
when the refrigerated product is in bottles, cans,
or other vapor-proof containers.

The most important factor governing the
humidity in the refrigerated space is the evapo-
ratorTD.* The smaller the difference in tempera-
ture between the evaporator and the space, the
higher is the relative humidity in the space.
Likewise, the greater the evaporator TD, the
lower is the relative humidity in the space.

When the product to be refrigerated is one
that will be affected by the space humidity, an
evaporator TD that will provide the optimum
humidity conditions for the product should be
selected. In such cases, the evaporator TD is
the most important factor determining the
evaporator selection. The design evaporator
TD required for various space humidities is
given in Table 11-2 for both natural-convection
and forced-convection evaporators.

In applications where the space humidity is
of no importance, the factors governing evapo-
rator selection are: (1) system efficiency and
economy of operation, (2) the physical space
available for evaporator installation, and (3)
initial cost.

11-17. The Effect of Air Circulation on
Product Condition. As stated previously,
circulation of air in the refrigerated space is
essential to carry the heat from the product to
the evaporator. When air circulation is inade-

* Some of the other factors which influence the
space humidity are: air motion, system running
time, type of system control, amount of exposed
product surface, infiltration, outside air conditions,
etc.

EVAPORATORS 17?

quate, the capacity of the evaporator is de-
creased, the product is not cooled at a sufficient
rate, the growth of mold and bacteria is encour-
aged, and sliming occurs on some products. On
the other hand, too much air circulation can be
as detrimental as too little. When the circula-
tion of air, is too great, the rate of moisture
evaporation from the product surface increases
and excessive dehydration of the product
results. Excessive dehydration can be very
costly in that it causes deterioration in product
appearance and quality and shortens the life
of the product. Furthermore, the loss of weight
resulting from shrinkage and trimming is a
considerable factor in dealer profits and in the
price of perishable foods.

The desired rate of air circulation varies with
the different applications and depends upon the
space humidity, the type of product, and the
length of the storage period.

With respect to product condition, air circula-
tion and space humidity are closely associated.
Poor air circulation has the same effect on the
product as high humidity, whereas too much air
circulation produces the same effect as low
humidity. In many instances, it is difficult to
determine whether product deterioration in
a particular application is caused by faulty air
circulation or poor humidity conditions. For
the most part, product condition depends upon
the combined effects of humidity and air
circulation, rather than upon the effect of either
one alone, and either of these two factors can
be varied somewhat, provided that the other is
varied in an off-setting direction. For example,
higher than normal air velocities can be used
without damage to the product when the space
humidity is also maintained at a higher level.

The type of product and the amount of

35
£ 30

a§2 5

l' E 15

uj 5

10*

15* 20* 25* 30* 35*
Evaporator TD

Fig. 11-21. Variation in evaporator capacity with
evaporator TD.

180 PRINCIPLES OF REFRIGERATION

exposed surface should be given consideration
when determining the desired rate of air circula-
tion. Some products, such as flowers and
vegetables, are more easily damaged by excessive
air circulation than others and must be given
special consideration. Cut meats, since they
have more exposed surface, are more susceptible
to loss of weight and deterioration than are
beef quarters or sides, and air velocities should
be lower. On the other hand, where the product
is in vapor-proof containers, it will not be af-
fected by high velocities and the rate of air
circulation should be maintained at a high level
to obtain the maximum cooling effect.

Recommended air velocities for product
storage are given in Tables 10-10 through 10-13.
11-18. Natural Convection Evaporators.
Natural convection evaporators are frequently
used in applications where low air velocities
and minimum dehydration of the product are
desired. Typical installations are household
refrigerators, display cases, walk-in coolers,
reach-in refrigerators, and large storage rooms.

The circulation of air over the cooling coil by
natural convection is a function of the tempera-
ture differential between the evaporator and the
space. The greater the difference in temperature,
the higher the rate of air circulation.

The circulation of air by natural convection
is greatly influenced by the shape, size, and
location of the evaporator, the use of baffles,
and the placement of the stored product in the
refrigerated space. Generally, shallow coils
(one or two rows deep) extending the entire

length of the cooler and covering the greater
portion of the ceiling area are best. As the
depth of the coil is increased, the coil offers
greater resistance to the free circulation of air
and the METD is thereby decreased with a
resulting decrease in the coil capacity. Since
cold air is denser than warm air and tends to
fall to the floor, evaporators should be located
as high above the floor as possible, but care
should be taken to leave sufficient room between
the evaporator and the ceiling to permit the free
circulation of air over the top of the coil.

For coolers less than 8 ft wide, single, ceiling-
mounted evaporators are frequently used. When
the width of the cooler exceeds 8 ft, two or more
evaporators should be used. In coolers where
there is not sufficient head room to permit the
use of overhead coils, side-wall evaporators
may be used. If properly installed, these will
function with approximately the same efficiency
as overhead coils. Typical overhead and side-
wall installations for large storage rooms are
shown in Figs. 11-22 and 11-23, respectively.

In small coolers, baffles are used with natural
convection coils to assure good air circulation.
The baffles are installed in such a manner that
they aid and direct the free flow of air over the
coil and throughout the refrigerated space. The
cold and warm air flues should each have an
area approximately equal to one-sixth or one-
seventh of the floor area of the cooler. Assum-
ing that the flues extend the full length of the
fixture, the width of the flues will then be
proportional to the width of the cooler. Since

Fig. 11-22. Overhead installa-
tion of natural convection
evaporator. Evaporator has
cast aluminum fins. (Courtesy
Detroit Ice Machine Company.)

EVAPORATORS

III

Ftf. II.23L Sid* will Installa-
tion of naturil tonvtalon
enpantor. (County On role
Ic* Miehmt Company )

warm Air has a greater specific volume than
eoJd air. some manufacturers recommend thai
the warm air flue tx a little larger than the cold
air flue. In Fig. 1 1-24, the width or the cold air
Due it equal to W/7, whereas the width of the
warm air flue is equal to W','6. The distance M)
from the coil to the ceiling should be approxi-
maldy equal to the width of the watm air flue
and never less than 3 in. Vertical side baffles
should extend approximately t in. above and

Hf. 11-24. Tvpkal baflla amngamaflt for natunl
conractlon end I.

3 (o 4 in. beto* the coil. The horizontal baffles
or coil decks should slope ) to 2 in. per foot to
give direction to the cold ajr flow and lo drain
on" the condensate- Also, I he coil decks must
be insulated so that moisture does not condone
on the undersurfcee of the deck and drip off on
the product. The dimension (if) is 4 to 7 in.,
whereas (O is usually 1 to 4 in,
IMf-Coi1-ajid-Bafr1eAii*mbll«s. The avail-
ability of radory-buill coJI-and-bafrk assemblies
has practically eliminated the need for the
custom building of baffles on the job. A typkal
ready built coil-and-bafflc assembly is shown
in Fig. 11-25. Since these assemblies are
available in a wide variety or sizes and combina-
tion (an Table R-l), they can be readily
applied to almost any natural convection
application.

11-20. Hating and Select I on uf Natural Con-
vection Evaporators, Basic capacity ratings
Tor natural convection evaporators, both prime
surface and finned, arc normally given in Btu/hr/
* F TD. However, in some instances, where it
will simplify evaporator selection, capacity
ratings are given for TD's other than I • F.

For the coil -and- baffle assemblies mentioned
in the preceding section, the ratings given arc
per inch of finned length. For bare pipe evapo-
rators, the ratings given are per square foot of
externa] pipe surface, although in some instances
bare pipe evaporators are rated per lineal foot
of pipe.

Ratings for plate evaporators are given per
squire foot of plate surface. Both sides of the
plate arc considered when computing the area

I K. MUNCltt.ES OF REfRlGEIUTlON

Ft|. 11*35, Nitunl «cn»«.

(Court -rty Dunhjm- faith. Int.)

of the plate, Frequently, ratings Tor plate
evaporators apply to an entire plate or to a
specific group or combination or plats,

Tvp^-.ii eating data f^? hhjcth typp of
■mural convection evaporators are given in
Tables R-l through R-7. The use or these
rating data in the selection of the various types
of evaporators is best illustrated through the
use or a series of examples.

Enamels J M. Select i natural convection
cou-and-baflfe assembly (Fig. 1 J -25 and Table
R-l) for the vegetable storage cool? in Example
KM I.

Solution. Since the capacity ratings for this
type of evaporator are given in Btufhr/T TD/in.
of finned length, the required evaporator capa-
city must be reduced to this value before a
selection can be made from the rating (able,
Abo. recall that a natural convection evapora-
tor should extend almost the full length of the
cooler in order to assure adequate air circulation
around the product.

From Example 10-11.
inside dimension of cooler

Required evaporator
capacity (average hourly
cooling load)

From Table 10-11. de-
sired space humidity for
mixed vegetable storage

From Table 1 1-2, design
evaporator TO required
forS7%RH

— 17 ft x9A

— 8500Biu/hr
*• Approx. 87%

To determine approxi-
mate finned length required :
Over-all length of
cooler (inside) — 17 ft

Allowing I ft on each
end of evaporator for
working space, the
approximate over-all
length of the evapo-
ratori4(«cFig.Jl-16)
(I7fl-2ft) -|5ft

According to the
manufacturer's speci-
fications {Table ft-1).
the over-all length of
the evaporator is 7 in
longer than the actual
finned length. Hence,
the approximate

finned length desired - 14 ft 3 in.
is {15 ft -7 in.) or 171 in.

To determine the require! capacity in Biu/hr/
*FTD,in. finned length:
Required evaporator
capacity per * F TD

Total c* J|H>rutor capacity

DewgiiL...!]«ralorTD
8500 Blu hr
14* Fin
- e0 l 7B•u;hr.■ , FTD
Required capacity
{Biu/hr/*FTD/ineh

m Rcquirett < o pacity per *FTD

Dei i rcdTtnned length
*07 Biu.hr/" F TD

171 in. tinned length
- 3.35 Btu/hrf F/ineh

EVAPORATORS IS)

Became or the width of the cooler, a two-
tcciicti evaporator will give the best results.
Reference to Tabic R-l indicates that Model
#PK-I6 with two fin* per inch (J -in. fin spacing)
has ■ capacity ofl 65 Biu/hr/ 1 F/in.

Using this mode) evaporator, the required
finned length is

J,*5 BtWnr/' F/in.

- lAfun

The overall length of the evaporator is 173
in. (166 in. + 7 in.) and. since the overall
length of the cooler inside is 204 in., the
dc*rancc between the evaporator and the cooler
wall at each end is 15 5 in.

The width of the evaporator should always
be checked a gains I the width of the cooler lo be
sure that the evaporator can be installed in the
space in accordance with the manufacturer's
recommendations. For evaporators of this
type, the manufacturer recommends that the
side of the evaporator be not less than 6 in. nor
more than 12 in. from the cooler wall {installa-
tion dimension A of Table R-l) and that the
distance between the two sections of the evapo-
rator (dimension C) be not less than 6 in. nor
more than 8 ft.

The maximum allowable evaporator width
can be determined by subtracting the minimum
of dimensions A and C from the inside width of
the cooler, viz :

Maximum width of evaporator •- Inside
width of cooler - {A + C + Q. In this par-
ticular case, the maximum allowable width of the
evaporator is

108 in. - (6 in. +6 in. + 6 in.) - 90 in.

Since the combined width of the two sections
of evaporator. Model #PK-16, is only 36 in.

fit. W-tt , Arnnf«rn«ni of natural (on**ct»on
twtporttori in itortj* COoUr (i*» Example 1 1. 1).

(IS in. x 2), the evaporator is suitable for the
cooler. A logical arrangement of ihc two
evaporator sections in the cooler is shown in
Fig 11-27.

To order the evaporator, specify the modet
number, fin spacing, and finned length, viz:
Model #PK- 1 6.3-166 in

Note. To avoid excessi vely long evaporators,
which are inconvenient to ship and install, a
multiple of evaporators should be used in large
coolers. Typical arrangements Tor large coolers
are shown in Fig. 11-28. These arrangements
are also suitable For plate banks.

ExampJ* 11*2, Using Table R-3, select
plate evaporators (banks) for ceiling installation
in the locker room of Example 10-16, To assure
good air circulation in the locker room, select
evaporators for a 15* FTD,

Solution, Analysis of room dimensions
(30.5 ft x 16 ft) indicates that four to six evapo-
rators (two or three banks installed end-to-
end over each aisle will be needed to provide
good ceiling coverage and adequate air distri-
bution. Reference to Table R-3 will show that
plate banks are available in stock lengths of
108 in. (9ft) and 144 in, (lift). Three banks
108 in. long (a total of 27 ft) or two banks 144
in, long (a total of 24 ft) could be installed
end-to-end over each aisle and allow adequate
working clearance at the ends.

Since the banks are already rated at the design
TD of 15* F, the ratings can be used directly and
the required capacity per bank can be deter-
mined by dividing the total hourly cooling load
by the desired number of plate banks, viz:

Hf> 1 1*1*, Arr»mj«m«ni of mmuni eonvteiion Required capacity _ Total cooling load (Btu/hr)
tviponton In ttorij* coolir («■ Exampi* 1 t-|). per bank (Btu/hr) Number of banks desired

inftdt hngei of coota?— I r ■

IB4 PRINCIPLES OF REFRIGERATION

In this instance, the capacity required per bank
is

24,650 Btu/hr
6 buna

24,650 Btu/hr
4 banks

-4lWBtWhr/13*FTD

= 6l62Blu/hr/)5' FTD

Referring to Table R-3, we see thai plate
bank. Model #5-1 210B-B, has a capacity of 4320
Btu/hr at a 15" F TD when operating below
32° F (frosted). This will permit good coverage
of the ceiling and at the same time allow suffi-
cient working space ui the ends of the banks (see
Fig- ll-29>.

In ordering the evaporators, specify refrig-
cram and the type of connections desired
(series or manifold), viz :

Mode) #6-J2l44-B, series connected for
Refrigerant- 12. Notice {Table R-J) that the
manufacturer specifies that two refrigerant flow
controls should be used with each bank for
Refrigeran 1-12, whereas only one a needed when
ammonia is the refrigerant.

Example 11-3, From Table R-4, select a
plate-stand assembly for the freezing cabinet in
Example 10-16. The inside dimensions of (be
cabinet are 28 in. x 90 in. and the freezing load
is 4590 Btu/hr. Base plate selection on 10 F
TD.

Fig. 11-26. Typical arrange rrmnfi tor natural con-
vection evaporator* In lirjt cooten.

Solution. Inspection of Tabte R-4 will show
that the ratings of the p I li : ._- stand assemblies
an based on a J 5° F TD. Since the design TD
in this instance is only 10 F. it is necessary to
determine (he capacity the plate must have at a
IV I TD in order to have the desired design
TD of 10- F. This is accomplished by dividing
the average hourly load by the design TD of
10* F and then multiplying by the rating TD of
15* F, vis;

4590 Btu/hr x 13' F
10" F

6885 Btu/hr

Thus, it is determined that an evaporator having
a capacity of 6885 Btu/hr ai a 15 ' F TD will
have I he desired capacity of 4590 Btu/hr at the
design TD of ID* F. Th* value can then be
used to select the plate stand directly from the
rating table.

By referring to Tabic R-4, plate stand,
Model #1-72%*-%. which is approximately 26 in.
wide and 88 in. long (including piping connec-
tions), will (it the freeaorr cabinet and has a
capacity of 7140 Btu/hr ai a 15" F TD (4760
Btu/hr at 10* F TD>, This provides a small
safety factor and is therefore satisfactory for
the application.

Alternate Solution. A rule of thumb used in
selecting plate freezer? for locker plant appli-
cations is to allow 0.5 Sq ft of plate surface for
each locker.

Allowing 0.5 sq ft of plate surface per locker
per day, the plate surface required la this
instance is

353 lockers x 0.5 = 1 7fi.S sq ft

By referring to Table R-2. the plate which best
flu the cabinet. Model #22S4 (22 in. x 84 in.),
has a surface area (both sides of plate) of 27.24
sq ft per plate. Therefore, seven pistes of
this size are required. Seven plates have a total
capacity of 7140 Btu/hr ai a I5 5 F TD or 4760
Btu/hr at a 10° F TD.

Example 11-4. Assuming a space tem-
perature of 0* F and a refrigerant temperature
of - 1 T F (1 7" F evaporalor TD), determine the
lineal feet of 1 i in. iron pipt required to handle
the cooling load on the locker room in Example
10-16.

Solution, By referring to Table R*6, the
capacity per square foot of pipe (outside surface)
at the given conditions is 1.5 Btu/hr/ F TD,
To determine the square feet of pipe surface
required, divide the average hourly load by the

EVAPORATORS IIS

Flf. 1 1 -It, Inttillillon of pliie
tanks in locker ptint (sm
Eximpl* 11-3.)

— 305"

|£ —

*— ^r^

** r tl IT

1 ^^^^^^^^^ '

*-r

capacity per square foot per degree TD and by
die design TD, viz:

Opacity required (Btu/hr)
Pipe capacity (Btu/hr/iq ft/' F) x TD

— Pipe surface I sq ft)

In this instance, the square feet of pipe surface
required is

24,650 Btu/hr

1.3 x 17 ™>»M"

By referring to Table R-7, 2.1 lineal ft of I J in.
pipe equals I sq ft of external pipe surface.
Hence, the lineal feet of I J in- pipe required is

966 x Z.3 - 2220 ft

11*21- Forced Convection Evaporator*.

Forced convection evaporators, commonly
called "unit coolers" or "blower coils" in com-
mercial refrigeration, are essentially finned coils
in a metal housing and equipped with
or more fans to provide air circulation.
Some typical unit cookrs on shown in Fig.
11-30.

The total cooling capacity of any evaporator
is directly related to the air quantity (cfm)
circulated over the evaporator. The air quantity
required for a given evaporator capacity
is basically a function of two factors: (i) the
sensible heat ratio and (2} the drop in the
temperature of the air passing over the cvapo*
rator. vix:

Cfm -

Total capacity (Btu/hr) x sensible heat ratio
Temperature drop of the air x 1.08

(11-5)

The sensible heat ratio is (he ratio of the
sensible cooling capacity of the evaporator to
the total cooling capacity. When air is cooled

below its dew point temperature, both the tem-
perature and the moisture content of the air an
reduced (Chapter 5). The temperature reduc-
tion is the result of sensible cooling, whereas the
moisture removed is the result of latent coo ting.
Kence. for an evaporator having a total cooling
capacity of one ton (12,000 Btu/hr) and a
sensible heat ratio of 0.85. the sensible cooling
capacity of (he evaporator is 10,200 Btu/hr
<W% or 12,000). whereas the latent cooling
capacity is 1800 Btu/hr (12,000 - 10.200).
Naturally, the sensible heal ratio of any
evaporator will depend upon the conditions of
the application, the design of the evaporate r, and
the air quantity. An average sensible heat ratio
for unit coolers is approximately 0.83.

As a general rule, the air temperature drop
through a well -designed unit cooler is approxi-
mately one- half the difference between the space
temperature and the refrigerant temperature.
For example, for an evaporator TD of 10 F.
the air temperature drop through the unit
u.H)lcr should be .tppri^inuid', f I

The constant LOS in Equation 1 1-3 is a con-
version factor involving air density (0J5), air
specific heat (0.24 Btu/lb/° F), and minutes per
hour (*0).*

£x m en p I b 1 1 -5. 1 V (ermine the approximate
quantity of air {cfm) circulated over a unit
cooler having a capacity of 20.000 Btu/hr if the
sensible heat ratio is 0,85 and the design evapo-
rator TD is 1 3* F.

Solution. Apply- _ 30.000 Btu/hr x 0.83
ing Equation 1 1-5, 7.5 x 1.08

the air quantity - 2)00 cfm

Ai a general rule, air velocities across the face
of unit coolers are maintained between 300 and
• Sen Section 14-4.

186 PRINCIPLES OF REFRIGERATION

Flf. Il-M. Topical Link tooter deslgni. Noikl thlt eool*r d*ligni iri mch that the air It not dttcharjid
directly on i ho *to red product. (Counety Dunham-Bush, Inc.)

500 ft per minute (Tpm), Although higher
velocities will result in higher transfer coeffi-
cients, they are not usually practical since they
alio increase fan horsepower requirements.
When the fan horsepower is increased beyond a
certain point, the additional heal given off by
the fan motor resulting from the increase it)
horsepower will exceed the increase in unit

cooler capacity resulting from higher air
velocities. Hence, the net effect in such cases
it to decrease, rather tkm increase, the over-
all capacity of the unit cooler. Too, where
the air velocity exceeds 500 fpm there Is a
tendency for moisture to he blown from the
face of the coil into the space and onto the
product.

The iir velocity {fpm) over the evaporator ii
a function of the air quantity (din) and the face
ana of the evaporator (v\ ft), viz :

Air quantity (cfm)

Velocity (fptro -

(11-6)

Face area |sq f i)

Example 1 1 ■*. Determine the face am of
thcevaporaior in Example 1 1 -5 if i he face velocity
is to be maintained at J 50 fpm.

Solution. By rearranging and _ 2100 dm
applying Equation 11*6, the " 350 fpm
face ana — 6 iq ft

1 1-22- Rat i ng and Saf action of U n i t Coolers ,
Basic rating! for unil cooler* are given in Blu/hr/
* F TD. For convenience, sometimes ratings
are given for 10" F and li* F TDs. Aa in the
case Of natural convection evaporators, the
deafen TD for unit cooler* depends primarily
on the space humidity requirements. In general,
for any given space humidity, the design TD for
unit coolers is about 4* F lo ** F leas than those
used for natural convection evaporator* (we
Table 1 1-2).

Since the air quantity is usually fixed by (he
fan selection at the time of manufacture, realiz-
ation of the rated capacity will depend primarily
upon whether or not the coil is kepi reasonably
free of frost by adequate defrosting When the
space is maintained below 34' F, some means of
automatic defrosting musi be used (see Chapter
20).

EVArOWKTORS 117

Example I J -7. From Table R-8, select a
forced convection evaporator (unit cooler)
suitable for installation in a beer storage cooler
having a calculated heal load of 16,600 Btu/hr.
Since apace humidity is not a factor, use JO F
TD for high system efficiency.

Solution. From Table R-S, select unit cooler
Model #UO!80 having a capacity of 18,000
Btu/hr at a 10* F TD, Since the unit cooler fan
motor operates inside the refrigerated space, the
motor heat becomes a part of the space cooling
load and must be added to the load calculations.
From Table R-g, the heat given off by the Tan
motor is 24,000 Btu/24 hr, Since the fan oper-
ates continuously to provide air circulation in
the refrigerated space, while the average hourly
cooling load is based on a 16-hr running time,
the average Btu per hour load resulting from
the fan motor heat is

34,000 BtiV?4 hr
16 far

- 1 300 Btu/hr

Hence, when the fan motor beat is considered,
the average hourly cooling load for the beer
cooler becomes 18,100 Btu/hr (16,600 + 1500).
Since the unit cooler selected has ■ capacity of
IS.OOO Btu/hr, it will be adequate for the
application.

Suggested locations for unit coolers in walk-
in refrigerators are shown in Fig. 11-11,

ftg. IIO I. Su(|«ikwii for
location of unit coolin In
walk-in f*frS|tf*ton. (From
tti* Mft£ Dots ftcok, D**Jjn
Volume, 1957-M edition, re-
produced by ptrmmlon of Ux
American Society of Hmlnf.
RvtriftrMint; and Air-Condi*

tlMldf Enjin«rv)

IBS PRINCIPLES Of REFRIGERATION

11-13. Liquid Chilling Evaporators, As with
air-cooling evaporators, liquid chilling evapo-
rators vary in type and design according to the
type of duly for which they art intended. Five
general types of liquid chiller* are in common
use: (1) the double^pipc cooler. (2) the Baudelot
cooler . (3) the lank-type cooler. (4) theshell*and-
coil cooler, and (3) the shell-and-tube cooler.
In all cases, the factors which influence the
performance of liquid chillers are the same as
those which govern the performance of air-
cooling evaporators and all other heat transfer

headers and art connected together by re-
movable return bends (inset I, The advantages
claimed for ink unit are rigid construction, the
elimination of refrigerant [uints, and easy
accessibility of the inner tubes for cleaning.

Double-pipe coolers ma;, ix; operated cither
dry-expansion or floodci!. In either cue,
counterflowing of the fluids in the lubes pro*
duces a relatively high heai transfer coefficient.
However, ihis type of cooler hits the disadvan-
tage of requiring more space, particularly head
room, than some of the oiher cooler design*.

Current conservative design values of heat transfer coefficient C based on
outside surface for bare tube coolers, unless mentioned otherwise, are as
follows:

Flooded shell -and- tube cooler (water to ammonia or R-12)

Flooded ihell-and -finned tube high velocity R-12 water cooler

Flooded shell-and-tube cooler (brine to ammonia)

Flooded shell-and-tube cooler (brine to R-I2>

Dry-expansion shel)-«nd-tube cooler, R-12 in tubes, water in shell

Baudelot cooler. Hooded (ammonia or R-12 to water)

Bau deloi cooler, dry-expansion (ammonia to water)

Baudelot cooler, dry-expansion (R-12 to water)

Double-pipe cooler (water to ammonia)

Double-pipe cooler (brine to ammonia)

Shell -and -coil cooler (water to ammonia)

Shell-and-coil cooler (water to R-12)

Spray-type shell-and-tube water coolers (ammonia or R-12)

Tank-and-agitator, coil-type water cooler, ammonia, flooded

Tank-and-agiiator, coil-type waler cooler. R-12 Hooded

Tank, ammonia, brine cooling, coils between can in k* tank

Tank, high velocity raceway type, brine to ammonia

Fig. It-XL Ht« tniwftr coeftidants (or vsrfout types of liquid chlllan. (Rtpi-miad from JMJ-5*
ASflE Dots Book, by furmlulon of th* Amtfican SoclKy of H**tlnj. ft«lri{«rsttr-i, «nd AJr-CondltJoiUnt
En|in*tn,)

For this reason, the double-pipe cooler is used
only in some few special appliea lions. A number
have been used in the wine-nuking and brewing
industries to chill wine and wort, and in ihe
petroleum industry for the shilling of oils.
1 1 -IS. Baudelot Coolers- The Baudelot cooler
shown in Hg. I l-M contisu of a series of hori-
zontal pipes which are located One under the
□I her and are connected i^geiher to form a
refrigerant circuit or circuit*. For either dry-
expansion or flooded operation, the refrigerant
is circulated through the inside of the tubes
while the chilled liquid flows in a thin film over

Min

|ft«

150

100

j«

115

100

200

150

120

150

125

ISO

250

125

100

110

surfaces. Heat transfer coefficients for average
designs of tome of the various chiller types are
listed in the table in Fig. 1 1-32.
11-24. Double-Pipe Coolers, The double-pipe
cooler consists of two lubes SO arranged that
one tube is inside the other. The chilled fluid
flows in one direction through the inner tube
while the refrigerant flows in the opposite
direction through the annular space between the
inner and outer tubes. One design of a double-
pipe cookr is shown in Fig, 1 1-33. In this design
the outer tubes arc welded to vertical refrigerant
headers while ihe inner lubes pass through the

EVAPORATORS IB?

Fij. 11-11. Double pip* cootar. Removable mum b*ndi (right) ira ditlfnad to make tub* readily
accessible, for clunlnj (Courtey Viltar rltnuficturini Company.)

the outside. The liquid fowl down over the
tubes by gravity from a distributor foaled at
the top of the cooler and is collected in a trough
at the bottom. The fact thai the chilled liquid
it at atmospheric pressure and is open to the air
nukes the Baudelot cooler ideal for any liquid
chilling application when Aeration u a factor .
The Baudejm chiller has been widely used for the
cooling of milk, wine, and wort, and for the

chilling of water Tor carbonation in bottling
plan is. With this particular type of chiller it is
possible lo chili liquid to a temperature very
close to the freezing point without the danger
of damaging the equipment if occasional
freezing of the liquid occurs.

Another advantage of the Baudelot cooler,
and one which is shared by the double-pipe
cooler, is that the refrigerant circuit is readily

Flf- 1 1-14. Bmddoi tooltr
employ *d in milk-cool I nj
application. (Coortu/ Dola
Refriierailn g Company.)

190 PRINCIPLES OF REFRIGERATION
_ Warm fcquWI m

Flf. If-JS. Typlal construc-
tion of tnnk-cjfpe liquid Caol a r.

J?Bfnj(er*n1

fines

split into several parts, a circumstance which
permits precoohng of the chilled liquid with
cold wain: before the liquid enter* the direct-
expansion portion of the cooler (see Fig. 1 7-34>.
11-16. Tank-Type Cooler*. The lank-type
liquid chiller consuls essentially of A bane-lube
refrigerant coil installed in the center or si one
side of a large steel lank which contain} the
chilled liquid. Although! completely Autxnefgctl
in the chilled liquid, the refrigerant coil is
separated from the main body of the liquid by

a baffle arrangement. As shown in Fig. 11-35.
a motor driven agitator is unlived to circulate
the chilled liquid over the cooling coil at
relatively high velocity, usual h, between LOO and
1 50 ft per minute, the liquet being drawn in at
one end of the coil compart men I .ind discharged
at the other cod.

The spiral-shaped, bare-mbc coils mentioned
in Section 1 1*4 and the race t-jt -type coil illus-
trated in Fig. 1 1-36 arc two coO designs fre-
quently employed in tank-i>pe chillers. With

M. Flooded rac*w*f colt, (Courttiy Vllltr Hlftufatlu t\ n| Company,}

IVAPOMATOHS

191

either design the coils are operated flooded.
The lee-Cel shown in Fig. 11-11 is another
variation or the tank-type chiller.

Tank -type chiller* can be Applied to any
liquid-chilling application where sanitation is
not a primary factor, and arc widely used for
the chilling or water, brine, and other liquids
to be used as secondary refrigerants. Because

a welded steel shell (Fig. 1 1-37), As a genera]
rule, the chiller is operated dry -expansion with
the refrigerant in the coils and ihe chilled liquid
imhc*hell. In a few cases, the chiller is operated
flooded, in which esse the refrigerant is in the
shell and the chilled liquid passe* through the
tubes. The former arrangement has ihe ad van*
lage of providing a holdover capacity, thereby

liquxs line

ntVffMlM

tc uuen dfatn

Thtrmal eaptAiion vahw

Flj. I I.JT- Shrll-ind-cdl cootsr, {Courtis/ Acm* tndintriai.)

of their inherent holdover capacity, they are
particularly suitable for applications subject to
frequent and seven fluctuations in loading. In
such cases, a comparatively large chilled-liquid
storage tank is provided in order to minimize
the rise in the temperature of the chilled liquid
during periods of peak demand. The advantage
gained by precooling is often considerable in
cases where the liquid to be chilled enters the
cooler at relatively high temperatures.
11-27. She 1 1 -and -Coil Cooler*. The shell-
and-coil chiller is usually made up of one or
more spiral-shaped, bare-tube coils enclosed in

making this type of chiller ideal for small
applications having high but infrequent peak
toads, tl is used primarily for the chilling of
water for drinking and for other purposes where
sanitation is a prime factor, as in bakeries and
photographic laboratories.

When operated Hooded with the refrigerant
in the shell, this type of chiller becomes what
is commonly referred to as an "instantaneous"
liquid chiller. One of the disadvantages of this
arrangement is that there is no holdover
capacity. Since the liquid is not recirculated,
it must be chilled instantaneously as it passes

192 PRINCIPLES OF REFRIGERATION

Liquid Outlet -

MM *c r ^"t , £ t i , 'l

f*l

Bf> tl-3*. Typical iheilmd-tub* chilltn. (0) Hooded typ«- Tuba bundla li ramovablc |b) Dry-txpuuion
Eype (refrigerant in lubm). Not* taffllnf of Wiur circuit Tub* theati ire fixed. (C*ur(t*y WonMnftOII

ihrough the coils. Another disadvantage is that
Ihc danger of damaging the chiller in (he event
of freeze-up is greatly increased in any chiller
where the chilled liquid is circulated through
the coils or tubes, rather than over the outside of
the tubes. For this reason, chillers employing
this arrangement cannot be recommended for
any application where it is required to chill the
liquid below 38° F,

Instantaneous shell -and -coil chillers arc used
principally for the chilling of beer and oiher
beverages in "draw-bars," in which case ihc
beverage is usually precooled to some extent
before entering the chiller.
1 1-2& Shell-and-Tvbe Chillers, Shell-and-
tube chillers have a relatively high efficiency,
require a minimum of floor space nnd head
room, are easily maintained, and are readily
adaptable to almost any type of liquid-chilling
application. For these reasons, the shcll-and-

tube chiller it by far the mosi widely used type.
Although individual designs il filer somewhat,
depending upon Ihc refrigen-ini used and upon
whether the chiller is operated dry-expansion
or flooded, the shcli-and-tube chiller consists
essentially of ft cylindrical Keel shell in which a
number of Straight tubes are arranged in parallel
and held in place at the ends by tube sheets.
When the chiller is operated dry-expansion, the
refrigerant is expanded into the tubes while the
chilled liquid is circulated ilmmgh the shell
(Fig. 1 1 -38b). When the chiller is operated
flooded, the chilled liquid is circulated through
the 1 iilics .111 J the refrigerant ii contained in the
shell . the level of the liquid refrigerant in the
shell being maintained with some type of float
control {Fig. LlOBd), In either case, the chilled
liquid is circulated through the chiller and con-
necting piping by means of a liquid circulating
pump, usually of the centrifugal type.

EVAPORATORS t?J

Shell diameters for shell~and-tube chillers
range from approximately 6 to GO in,, and the
number of tubes in the shell varies from fewer
Lhan SO to several thousand- Tube dkmclem
range from I in. through 2 in., and lube lengths
vary from S to 20 ft. Chillers designed for use
with ammonia are equipped with steel tubes,
whereas those intended for use with other
refrigerant! an usually equipped with copper
tubes in order to obtain a higher heat transfer
ooefnaenl. Because of the rela lively low film
conductance of halocarbon refrigerants, chillers
designed for use with these refrigerants are often
equipped with tubes which are fumed on the
refrigerant side. In the case of dry-expansion
chillers, the lubes are finned internally with
longitudinal fins of the types shown in Fig.
11-12. For flooded operation, the tubes are
finned externally using a very short fin which
protrude* from the lube wall only approximately
,\thof an inch.

As a general rule, dry<*p*nlkm drillers are
employed in small and medium tonnage instal-
lations requiring capacities ranging from 2 to
approximately 250 tons, but ore available in
larger capacities. Flooded chillers, available
in capacities ranging from approximately 10
through several thousand tons, are more
frequently applied in ihe larger tonnage instal-
lations.

Il-M. Dry-Expansion CMEIar*. The principal
advantages of the dry-expansion chiller over the

flooded type me the smaller refrigerant charge
required and the assurance of positive oil return
to the compressor. Too, aa previously stated,
the possibility of damage to the chiller in the
event of freeze-up is always considerably less
when the chilled liquid is circulated over the
lubes ml her lhan through them. The more
important construction details of several designs
of dry-cupansion chillers are shown in Figs.
1 1-39 "and 1 1-40.

En order to maintain the liquid velocity
wilhin ihc limits which will produce the most
effective heat transfer-pressure drop ratio, ihe
vctochy of the chilled liquid circulated over the
lubes is controlled by varying the length and
spacing of the segmental baffles, When the
fiow rate and/or liquid viscosity is high, short,
widely spaced taffies are used to reduce the
velocity and minimize the pressure drop through
the chiller. When the flow rate and/or liquid
viscosity is low. longer, more closely spaced
baffles are used in order to increase fluid
velocity and improve the heat transfer coef-
ficient (Fig. IMkr).

The number and the length of the refrigerant
circuits required to maintain the refrigerant
velocity through the chiller tubes within reason-
able limits depend on the total chiller load and
on the relationship of the chilled liquid flow
rate to the METD. Since the* factors vary
with the individual application, it follows that
the optimum refrigerant circuit design atso

Flf . I !-«. C u«w*¥ Mrtloit llltMirstini eoimrucrion dtutli of drr-wtpUMlon chllfer with ftxad tub* ihMta,
(Counter Aem* Industrial.)

SM PRINCIPLES OF REFRIGERATION

Flf. I MO. Dry-expirtKon (hilJ*r with tub* bundl* fnnl*riy rtmovtd IO llMW lutw inwjimint mnif
refrlferani flUtribirtcn, Tub* bundl* w b* rtrnov*d U t unll, (CourtMjf Ktnmrd Cn.non, Amtritln Air
Fihmr Com piny , Inc.)

varies with the individual application. For this
reason, chillers arc made available with cither
single or multiple refrigerant circuits of varying
lengths. For the design shown in Tig. 11-39,
the number and length or the refrigerant circuits
depend on the lube length and on the arrange-
ment of the baffling in the end-plates or refriger-
ant beads which are bolted to the lube sheets
at the ends of the chilter. The refrigeranl
circuit arrangement for any one model chiller
can be changed by -changing the refrigerant
heads (Fig. 11 -41 A}'

11-30. Flooded Chilian, Standard flooded
chiller designs include both single and multipass
lube arrangements, lor single pass flow, the
tubes ere so arranged lhat ihe chilled liquid
passes through all the lubes simultaneously
and in only one direction.

Multipass circulation of the chilled liquid
through the ehilkr is accomplished through the
use of baffled end-plates or heads which are
boiled lo the ends or the chiller (Fig. 1 1-42),
The arrangement of the end-plate baffling
determines the number of passes the chilled
liquid makes from one to the other before
leaving the chiller. Although two, four, and

siX'put arrangements are ihe most common,
more pears are used in many instances.

As in the case of the dr\ -expansion chiller,
some Hooded chillers are deu^ned with remov-
able tube bundles, whereas other? have fixed
lube sheets. In the fixed tube sheet design, the
tube sheets are welded to the shell so that
the lube bundle is not removable. However,
by unbolting the end-plate* the tubes become
readily accessible for cleaning and individual
lubes can be removed and replaced if necessary.
The chillers shown in Fig* LI 3Bfr and 11-39
employ fixed tube sheets, whereas those in Figs.
M->3Bu and I MO have tube huiidlcs.

In some flooded chiller designs, the shell
is only partially filled with tubea in order to
provide a large vapor-disen^igmg area and rela-
tively low velocity in the spate above the tubes
(Fig. 11*43). Th n design dim i n.j | e<i the possibility
of liquid carry-over into the suction line and
therefore is particularly well mi i fed to sudden
heavy increases in loading

In those chiller designs where Ihe shell is
completely filled with tubes, a surge drum or
accumulator should be used in separate any
entrained liquid from the vapor before the vapor

EVAPORATORS IW

enter* the suction line. Some flooded chiller*
■re equipped with bull [-in liquid-suction heal
exchanger* (Fig. 11-45). Although I he prima ry
function of the heal exchanger is to insure that
oniy dry vapor enter* the suction line, it has
the additional effect of increasing the efficiency
of the chiller in that it subcools the liquid
approaching the chiller and thereby reduces the
amount of flash gas that enter* the cooler.

Short cut on-*** ipKir^g

taint* tm n**cs -intern*! vww

4-FHs

JOtiHh

a-ataa

2 Circuits

•^

Flf. I Ml. (o) Baffla ipiCinf in d ry-** pwulon ehil-
kfi. (D) Typ+oJ refrliertnt headl far dry-*Kp*nile>n
chiller. (CDurte»r of Acme InduitrkM.)

The vertical ihcll-and-lubc chiller shown in
Fig. 11-44 has the advantage of requiring a
minimum amount of floor space. The chiller
is operated flooded. The chilled liquid enters
the chiller at the top and flows by gravity down
the inside of the tubes. A circulating pump
draws chilled liquid from the storage tank at the
bottom and delivers it through the connecting
piping- The return liquid is piped to the dis-
tributor box at the top. from where it again
flown through the lubes. A specially designed
distributor installed at the lop of each tube
(inset) imparts a. swirling motion to the chilled
liquid, which causes the liquid to flow in a. com-
paratively thin film down the inside tube
turfites,

11-31. Spray-Type Chillers. The spray-type
chiller is similar in construction to the conven-
tional flooded chiller cuoep! that the liquid
refrigerant is sprayed over the outside of the
waier tubes from nozzle* located in a spray
header above the tube bundle (Fig. 20-19). The
unevaporated liquid drains from the tube into a
tump at the bottom of the chiller from where it
is recirculated to the spray nozzles by ■ low head
liquid pump. A high recirculation rate assures
continuous wetting of the tube surfaces and
results in a high rale of heat transfer.

The principal advantages of this type of
chiller is its high cflicicncy and relatively small
refrigerant charge. Disadvantages are the high
installation cost and the need for a liquid
recirculating pump.

1 1 ■&. Chiller Selection Procedure. Although
selection methods differ somewhat depending
upon the type of chiller and the particular manu-
facturer, all are based on the simple funda-
mentals of heat transfer and fluid now which
have already been described. Almost without
exception, manufacturers include sample selec-
tion procedure along with the design and
capacity data in their e quip ment catalogs. The
following selection procedure follows very
closely that given in the catalog of one
manufacturer for the selection of dry-expansion
chillers-*

Example II-*, ]i is desired to cool 50 gpm
of wrner from 54° F to 46° F with a refrigemnt
tempera lure ai measured at the cooler outlet of
40- F usinj; Refrigerant* 1 2.

* Acme Industrie*. Inc.

W PRINCIPLES OF REFRIGERATION

Fig. 11*41. Flooded thlller d**F(n«d far multlp»I circulation of chiliad liquid. I",>. pm circulation i*
ttcompliihed by mean* of th* C4ffl«d *fld-£l«*-, or writer heads which ir* bc-lud to Ihe endi of the chiller.

{CourtHr Vilter Minu^ctuhni Company.)

Solution

Step 1. Determine ihe total chiller load

tons.

produces the highest pumping had Hence, if
space it not ■ problem, ihe must logical choice
would lean to be type 8M I U-wcvcr, a check

Gpm x 500 x cooling range

1 2.000 Blu/hr/ton
SO x 500 x (54 - 46)

> 2,000

— 16.7 ions

Step 2. Determine the mean effective tem-
perature difference (METD),
Water in minus refrig-
erant temperature 54 — 40 ■■ 14* F LTD
Water out minus re-
frigerant temperature -Mi - 40 — 6" F STD
From Table II -1, METD - 9.47* F

Step 3. Select trial chiller {shell diameter and
baffles spacing) from Fig, I of Table R-9. Enter
Fig. 1 at 50 gpm on the lower vertical scale and
move horizontally across the chart to the
diagonal line representing the type unit desired,
The number indicates shell diameter and the
letter indicates baffle spicing, Possible choices
are 10M, I2L, SM, 12K, 10K, and SL, As a
general rule, small diameter chillers are more
economical, whereas large diameter chillers are
more compact. Type M burning produces the
lowest pumping head, whereas type L baffling

Fig. I ML Flooded chiller with shell only partially
filled with tubst In order to provide i lirgc vapor
d-Hniafln & tree ibovi the tubri. {CourtMy Worth-
ln|tdn Corporation.)

EVAPORATORS

197

will show that neither 8M nor 8L is available
with sufficient surface area in this instance.
Therefore, select type 10M (8 to 30 tons). From
the point of intersection move vertically upward
to a diagonal line in the upper portion of Fig. 1
which represents a METD of 9.47° F as found

Water
inlet

Gas outlet -

14 ft DXH chiller has a surface area of 184 sq ft
(Model No. DXH-1014).

Step 6. Determine the water pressure drop
through the chiller. From the bottom of Fig. 1,
Table R-9, the pressure drop per foot of length
with type M baffling is 0.425 ft of water column.

Refrigerant
feed

Inset

Fig. 11-44. Vertical shell-and-tube "Spira-Flo" chiller designed for flooded operation. The water flowing
down through the tubes is given a swirling action by specially designed nozzles (inset). (Courtesy Worthington
Corporation.)

in Step 2. From this intersection move hori-
zontally to the scale at the left margin and read
the loading of 1110 Btu/hr/sq ft (loading is the
U value times the METD).
Step 4. Determine the surface area required.

Surface area -^^
Loading

200.000

-^5-- 180.2 sq ft

Step 5. Select chiller length from Table R-9
to meet surface area requirements. A 10 in. x

Pressure drop = length (feet) x pressure drop/
foot

14 feet x 0.425 = 5.95 ft H a O

11-33. Direct and Indirect Systems. Any

heat transfer surface into which a volatile liquid
(refrigerant) is expanded and evaporated in
order to produce a cooling effect is called a
"direct-expansion" evaporator and the liquid
so evaporated is called a "direct-expansion"
refrigerant. A direct-expansion or "direct"

198 PRINCIPLES OF REFRIGERATION

Brine coil "V.

WWW/WW/WW/W/W,

Refrigerated

space or

material

V/s;///;ss///s;;j;ss;;ss;;;//v.

Liquid from
receiver

from /"
ver jfVl

* — &-i

Vapor to

compressor

suction

Warm brine_
to chiller "

Refrigerant control

I Brine solution!

?mrm/m/m///??r#mrA

Cold brine
to coil

Brine pump

Fig. I M5o. Indirect system.

■ Brine coil

t4-

Cold air to
refrigerated space

Warm brine
to chiller

receiver ,^^BZ^mB^.

Refrigerant control

'**r .

Vapor to rx. s;

compressor z |

suction vM

.. .

^ £ " w wiw hw i p wbh w/ ■* >7

Brine solution {

7~\

Cold brine
to coil

Warm air
from space

-Air duct

^ ^-Brine

-Brine pump
Fig. Il-45b. Indirect system — brine coil in communicating duct.

refrigerating system is one wherein the system
evaporator, employing a direct-expansion refrig-
erant, is in direct contact with the space or
material being refrigerated, or is located in air
ducts communicating with such spaces. Up to
this point, only direct refrigerating systems have
been considered.

Very often it is either inconvenient or un-
economical to circulate a direct-expansion
refrigerant to the area or areas where the cooling

is required. In such cases, an indirect refrigerat-
ing system is employed. Water or brine (or
some other suitable liquid) is chilled by a direct-
expansion refrigerant in a liquid chiller and
then pumped through appropriate piping to the
space or product being refrigerated. The
chilled liquid, called a secondary refrigerant,
may be circulated directly around the refriger-
ated product or vessel or it may be passed
through an air-cooling coil or some other type

EVAPORATORS

199

of heat transfer surface (Fig. 11-45). In either
case, the secondary refrigerant, warmed by the
absorption of heat from the refrigerated space
or product, is returned to the chiller to be
chilled and recirculated.

Indirect refrigerating systems are usually
employed to an advantage in any installation
where the space or product to be cooled is
located a considerable distance from the con-
densing equipment. The reason for this is that
long direct-expansion refrigerant lines are
seldom practical. In the first place, they are
expensive to install and they necessitate a large
refrigerant charge. Too, long refrigerant lines,
particularly long risers, create oil return prob-
lems and cause excessive refrigerant pressure
losses which tend to reauce the capacity and
efficiency of the system. Furthermore, leaks
are more serious and are much more likely to
occur in refrigerant piping than in water or
brine piping.

Indirect refrigeration is required also in many
industrial process cooling applications where
it is often impractical to maintain a vapor tight
seal around the product or vessel being cooled.
Too, indirect systems are used to an advantage
in any application where the leakage of refriger-
ant and/or oil from the lines may cause con-
tamination or other damage to a stored product.
The latter applies particularly to meat packing
plants and large cold storage applications when
ammonia is used as a refrigerant.
1 1-34. Secondary Refrigerants. Some com-
monly used secondary refrigerants are water,
calcium chloride and sodium chloride brines,
ethylene and propylene glycols, Methanol
(methyl alcohol), and glycerin.

Almost without exception, water is used as
the secondary refrigerant in large air condition-
ing systems and also in industrial process cooling
installations where the temperatures maintained
are above the freezing point of water. Water,
because of its fluidity, high specific heat value,
and high film coefficient, is an excellent
secondary refrigerant. It also has the advantage
of being inexpensive and relatively noncorrosive.
In air conditioning applications, the chilled
water is circulated through an air cooling coil
or through a water spray unit. In either case,
the air is both cooled and dehumidified. In the
water spray unit, the water i. sprayed from
nozzles and collected in a pan or basin at the

bottom of the spray unit, from where it is
returned to the chiller. Since the air passing
through the water spray is chilled below its dew
point temperature, a certain amount of water
vapor is condensed from the air and is carried
to the basin with the spray water. With either
the cooling coil or the spray unit, the amount
of cooling and dehumidification can be con-
trolled by varying the amount and temperature
of the chilled water.

Water is also used frequently as a secondary
refrigerant in small beverage coolers and in
farm coolers designed for cooling milk cans.
In such cases, the water, because of its high
conductivity, permits more rapid chilling of the
product than would be possible with air. Too,
the water supplies a holdover capacity which
tends to level out load fluctuations resulting
from intermittent loading of the cooler.
11-35. Brines. Obviously, water cannot be
employed as a secondary refrigerant in any
application where the temperature to be main-
tained is below the freezing point of water. In
such cases, a brine solution is often used.

Brine is the name given to the solution which
results when various salts are dissolved in water.
If a salt is dissolved in water, the freezing tem-
perature of the resulting brine will be below the
freezing temperature of pure water. Up to a
certain point, the more salt dissolved in the
solution, the lower will be the freezing tempera-
ture of the brine. However, if the salt concen-
tration is increased beyond a certain point, the
freezing temperature of the brine will be raised
rather than lowered. Hence, a solution of any
salt in water has a certain concentration at
which the freezing point of the solution is lowest.
A solution at the critical concentration is called
a eutectic solution. At any concentration above
or below this critical concentration, the freezing
temperature of the solution will be higher, that
is, above the eutectic temperature.* When the
salt content of the brine is less than that which
is required for a eutectic solution, the excess
water will begin to precipitate from the solution
in the form of ice crystals at some temperature
above the eutectic temperature. The exact tem-
perature at which the ice crystals will begin to

* At any temperature other than the eutectic
temperature, the term "freezing temperature" is
used to mean the temperature at which ice or salt
crystals begin to precipitate from the solution.

200 PRINCIPLES OF REFRIGERATION

form depends upon the degree of the salt con-
centration and upon the relative solubility of the
salt in water, the latter factor decreasing as
the temperature of the solution decreases. The
continued precipitation of ice crystals from the
solution as the temperature is reduced causes a
progressive increase in the concentration of the
remaining brine until, at the eutectic tempera-
ture, a slush consisting of ice and eutectic brine
will exist. The further removal of heat from this
mixture will result in solidification of the eutec-
tic brine. Solidification of the eutectic brine will
take place at a constant temperature.

On the other hand, when the salt content of
the brine is in excess of the amount required for
a eutectic solution, the excess salt will begin to
precipitate from the solution in the form of salt
crystals at some temperature above the eutectic
temperature. Continued precipitation of salt
from the mixture as the temperature is reduced
will result in a mixture of salt and eutectic brine
when the eutectic temperature is reached. The
further removal of heat from the mixture will
result in solidification of the eutectic brine at
constant temperature.

Two types of brine are commonly used in
refrigeration practice: (1) calcium chloride and

Air -

Air

iff Mr M

ii '"

Brine from
chiller

Brine to
concentrator

Concentrator

I Brine

\, Brine from
concentrator

*" Brine to
chiller

Fig. 1 1-46. Brine spray cooler.

(2) sodium chloride. The two brines are pre-
pared from calcium chloride (CaCl^ and sodium
chloride (NaCl) salts, respectively, the latter
salt being the common table variety.

Calcium chloride brine is used primarily in
industrial process cooling, in product freezing
and storage, and in other brine applications
where temperatures below 0° F are required.
The lowest freezing temperature which can be
obtained with calcium chloride brine (the eutec-
tic temperature) is approximately —67° F. The
salt concentration in the eutectic solution is
approximately 30% by weight. The freezing
temperature of various concentrations of cal-
cium chloride brine are given in Table 11-3,
along with some of the other important pro-
perties of the brine.

The principal disadvantage of calcium
chloride brine is its dehydrating effect and its
tendency to impart a bitter taste to food pro-
ducts with which it comes in contact. For this
reason, when calcium chloride brine is used in
food freezing applications, the system must be
designed so as to prevent the brine from coming
into contact with the refrigerated product.

Sodium chloride brine is employed mainly
in those applications where the possibility of
product contamination prevents the use of
calcium chloride brine. Sodium chloride brine
is employed extensively in installations where
the chilling and freezing of meat, fish, and other
products are accomplished by means of a brine
spray or fog.

The lowest temperature obtainable with
sodium chloride brine is approximately —6° F.
For this freezing temperature the salt concen-
tration in the solution is approximately 23%.
The thermal properties of sodium chloride brine
at various concentrations is given in Table 11-4.

It is of interest to notice that the thermal
properties of both calcium chloride and sodium
chloride brines are somewhat less satisfactory
than those of water. As the salt content of the
brines is increased, the fluidity, specific heat
value, and thermal conductance of the brines all
decrease. Hence, the stronger the brine solu-
tion, the greater the quantity of brine that must
be circulated in order to produce a given
refrigerating effect.

Since the specific gravity of the brine increases
as the salt concentration increases, the degree of
salt concentration and the thermal properties of

EVAPORATORS 201

Air -*■

Fig. 1 1-47. Brine spray cooler. Brine WW-

Direct
expansion ■
evaporator

M Eliminators 1

■■MhHAai.'i

Brine to
concentrator

To condensing
unit

the brine can be determined by measuring the
specific gravity of the brine with a hydrometer.
11-36. Antifreeze Solutions. Certain water
soluble compounds, generally described as anti-
freeze agents, are often used to depress the
freezing point of water. The more widely known
antifreeze agents are ethylene glycol, propylene
glycol, Methanol (methyl alcohol), and glycerin.
All these compounds are soluble in water in all
proportions. The freezing temperature of water
in solution with various percentages of each of
these compounds is given in Table 11-5.

Propylene glycol is probably the most exten-
sively used antifreeze agent in refrigeration
service. In common with ethylene glycol,
propylene glycol has a number of desirable
properties. Unlike brine, glycol solutions are
noncorrosive. They are also nonelectrolytic and
therefore may be employed in systems containing
dissimilar metals. Being extremely stable com-
pounds, glycols will not evaporate under normal
operating conditions. Because of the many

advantages of glycol solutions, they are being
used to replace brines in a number of installa-
tions, particularly in the brewing and dairy
industries. The change-over from brine to glycol
can be accomplished with practically no change
in the plant facilities.

1 1-37. Brine Spray Units. Like chilled water,
the chilled brine (or antifreeze solution) may be
circulated directly around the refrigerated pro-
duct or container, or it may be used to cool the
air in a •refrigerated space. When used to cool
air, the chilled brine is circulated through a
serpentine coil or through a brine spray unit.
Two types of brine spray units which have been
used extensively are shown in Figs. 11-46 and
11-47. In the former unit chilled brine from a
brine chiller located outside the refrigerated
space is- sprayed down from spray nozzles and
collected in the basin of the unit, from where it
is returned to the brine chiller. In the latter type
the brine is chilled by a direct-expansion coil
located within the brine spray unit itself.

202 PRINCIPLES OF REFRIGERATION

PROBLEMS

1. A walk-in cooler 8 ft by 9 ft by 9 ft high has
walls 6 in. thick and is maintained at a tempera-
ture of 35° F. The load on the cooler is 7500
Btu/hr.

(a) Select a natural convection cooling coil
(Plasti-Cooler) which will produce a
relative humidity of approximately 85%
in the cooler.

(6) Select a Unit Cooler which will produce
approximately the same conditions in the
cooler.

2. A freezing cabinet 6 ft high, 25 in. deep, and
80 in. wide has a freezing load of 3600 Btu/hr.

Based on an evaporator TD of 10° F, determine
the size and number of individual freezer plates
to be used as shelves in the freezing cabinet.

3. The load on a tank-type brine cooler is
4500 Btu/hr. The brine is to be maintained at
a temperature of 35° F with a refrigerant tem-
perature of 19° F. Assuming little or no
agitation of the brine, determine the lineal feet
of | in. pipe required for the evaporator.

4. It is desired to cool 100 gpm of water from
56° F to 46° F with a refrigerant temperature
38° F at the cooler outlet. Select an appro-
priate chiller and determine the water pressure
drop through the chiller in psi.

Performance
of Reciprocating
Compressors

12-1. Refrigeration Compressors. Vapor
compressors used in refrigeration are of three
principal types: (1) reciprocating, (2) rotary,
and (3) centrifugal. Of the three, the recipro-
cating compressor is by far the" one most
frequently used.

Rotary compressors are limited to use in very
small fractional horsepower applications, such
as home refrigerators and freezers and small
commercial applications. Even in this limited
area, rotary compressors represent only a small
fraction of the total number used. Some rotary
compressors are used also as booster compres-
sors.* Their use for this purpose appears to be
increasing.

Centrifugal compressors are used only on
very large applications, usually at least SO tons
or above. In this area, they are widely accepted
and are rapidly increasing in number because
the number of large applications is growing
steadily.

Only the performance of reciprocating com-
pressors will be discussed in this chapter.
Reciprocating compressor design, along with
the design and performance of rotary and centri-
fugal compressors, is discussed elsewhere in the
text at a more appropriate time and place.
However, much that is said in this chapter about

* Booster compressors are discussed in Chapter

20.

the performance of reciprocating compressors
will apply also to the performance of rotary and
centrifugal compressors.
12-2. The Compression Cycle. Before at-
tempting to analyze the performance of the
compressor, it is necessary to become familiar
with the series of processes which make up
the compression cycle of a reciprocating com-
pressor.

A compressor, with the piston shown at four
points in its travel in the cylinder, is illustrated
in Fig. 12-1. As the piston moves downward
on the suction stroke, low-pressure vapor from
the suction line is drawn into the cylinder
through the suction valves. On the upstroke of
the piston, the low-pressure vapor is first com-
pressed and then discharged as a high-pressure
vapor through the discharge valves into the head
of the compressor.

To prevent the piston from striking the valve
plate, all reciprocating compressors are designed
with a small amount of clearance between the
top of the piston and the valve plate when the
piston is at the top of its stroke. The volume of
this clearance space is called the clearance
volume and is the volume of the cylinder when
the piston is at top dead center.

Not all the high-pressure vapor will pass out
through the discharge valves at the end of the
compression stroke. A certain amount will
remain in the cylinder in the clearance space
between the piston and the valve plate. The
vapor which remains in the clearance space at
the end of each discharge stroke is called the
clearance vapor.

Reference to Figs. 12-2 and 12-3 will help to
clarify the operation of the compressor. Figure
1 2-2 is a time-pressure diagram in which cylinder
pressure is plotted against crank position.
Figure 12-3 is a theoretical pressure- volume
diagram of a typical compression cycle. The
lettered points on theTP andPV diagrams corre-
spond to the piston positions as shown in
Fig. 12-1.

At point A, the piston is at the top of its
stroke, which is known as top dead center. When
the piston is at this position, both the suction
and discharge valves are closed. The high
pressure of the vapor trapped in the clearance
space acts upward on the suction valves and
holds them closed against the pressure of the
suction vapor in the suction line. Because the

203

204 PRINCIPLES OF REFRIGERATION

pressure of the vapor in the head of the com-
pressor is approximately the same as that of the
vapor in the clearance volume, the discharge
valves are held closed either by their own weight
or by light spring loading.

As the piston moves downward on the suction
stroke, the high-pressure vapor trapped in the
clearance space is allowed to expand. The

Discharge^

180
Crank position

Fig. 12-2. Theoretical time-pressure diagram of
compression cycle in which cylinder pressure is
plotted against crank position.

piston reaches the bottom of its stroke at point
C. During the time that the piston is moving
from B to C, the cylinder is filled with suction
vapor and the pressure in the cylinder remains
constant at the suction pressure. At point C,
the suction valves close, usually by spring action,
and the compression stroke begins.

The pressure of the vapor in the cylinder
increases along line C-D as the piston moves
upward on the compression stroke. By the time
the piston reaches point D, the pressure of the
vapor in the cylinder has been increased until
it is higher than the pressure of the vapor in the
head of the compressor and the discharge valves
are forced open; whereupon the high-pressure
vapor passes from the cylinder into the hot gas
line through the discharge valves. The flow of

Fig. 12-1. (a) Piston at top dead center, (b) Suction
valves open, (c) Piston at bottom dead center, (d)
Discharge valves open.

expansion takes place along line A-B so that
the pressure in the cylinder decreases as the
volume of the clearance vapor increases. When
the piston reaches point B, the pressure of the
re-expanded clearance vapor in the cylinder
becomes slightly less than the pressure of the
vapor in the suction line; whereupon the suction
valves are forced open by the higher pressure in
the suction line and vapor from the suction line
flows into the cylinder. The flow of suction
vapor into the cylinder begins when the suction
valves open at point B and continues until the

Clearance

Volume of

re-expanded •

clearance vapor

Volume

Fig. 12-3. Pressure-volume diagram of typical com-
pression cycle.

PERFORMANCE OF RECIPROCATING COMPRESSORS 205

the vapor through the discharge valves con-
tinues as the piston moves from D to A while
the pressure in the cylinder remains constant at
the discharge pressure. When the piston returns
to point A, the compression cycle is completed
and the crankshaft of the compressor has
rotated one complete revolution.
12-3. Piston Displacement. The piston dis-
placement of a reciprocating compressor is the
total cylinder volume swept through by the
piston in any certain time interval and is usually
expressed in cubic feet per minute. For any
single-acting, reciprocating compressor, the pis-
ton displacement is computed as follows:

v,--

n-Z)* x L x N x n
4 x 1728

(12-1)

where V„ -» the piston displacement in cubic

feet per minute
D = the diameter of the cylinder (bore)

in inches
L = the length of stroke in inches
N = revolutions of the crankshaft per

minute (rpm)
n — the number of cylinders

The volume of the cylinder which is swept
through by the piston each stroke (each revolu-
tion of the crankshaft) is the difference between
the volume of the cylinder when the piston is
at the bottom of its stroke and the volume of
the cylinder when the piston is at the top of its
stroke. This part of the cylinder volume is
found by multiplying the cross-sectional area of
the bore by the length of stroke. Thus:

Cross-sectional area of the
bore in square inches

Volume of cylinder swept
through by the piston each
stroke in cubic inches

ttD*

Once the cylinder volume is known, the total
cylinder volume swept through by the piston of
a single cylinder compressor each minute in
cubic inches can be determined by multiplying
the cylinder volume by the rpm (N). When the
compressor has more than one cylinder, the
cylinder volume must also be multiplied by the
number of cylinders («). In either case, dividing
the result by 1728 will give the piston displace-
ment in cubic feet per minute.

Example 12-1. Calculate the piston dis-
placement of a two cylinder compressor
rotating at 1450 rpm, if the diameter of the
cylinder is 2.S in. and the length of stroke is 2
in.

Solution. Substituting in Equation 12-1,
3.1416 x (2.5)* x 2 x 1455 x 2
4 x 1728

= 16.52 cu ft/min

12-4. Theoretical Refrigerating Capacity.

The refrigerating capacity of any compressor
depends upon the operating conditions' of the
system and, like system capacity, is determined
by the weight of refrigerant circulated per unit
of time and by the refrigerating effect of each
pound circulated.*

The weight of refrigerant circulated per
minute by the compressor is equal to the weight
of the suction vapor that the compressor
compresses per minute. If it is assumed that the
compressor is 100% efficient and that the
cylinder of the compressor fills completely with
suction vapor at each downstroke of the piston,
the volume of suction vapor drawn into the
compressor cylinder and compressed per minute
will be exactly equal to the piston displacement
of the compressor. The weight of this volume
of vapor, which is the weight of refrigerant
circulated per minute, can be calculated by
multiplying the piston displacement of the
compressor by the density of the suction vapor
at the compressor inlet.

Once the weight of refrigerant compressed
per minute by the compressor has been deter-
mined, the theoretical refrigerating capacity of
the compressor in tons can be found by multi-
plying the weight of refrigerant compressed per
minute by the refrigerating effect per pound and
then dividing by 200.

Example 12-2. The compressor in Example
12-1 is operating on a R-12 system at a suction
temperature of 20° F. If the suction vapor
reaching the compressor inlet is saturated and
if the temperature of the liquid at the refrigerant
control is 100° F, determine

(a) the total weight of refrigerant circulated
per minute

(6) the theoretical refrigerating capacity of
the compressor in tons.

* Since it is the compressor which circulates the
refrigerant through the system, compressor capacity
and system capacity are one and the same.

206 PRINCIPLES OF REFRIGERATION

Solution

(a) From Example 12-1,
piston displacement
From Table 16-3, den-
sity of R- 12 saturated
vapor at 20° F

Weight of refrigerant
circulated per minute

(b) From Table 16-3,
enthalpy of R-12 satur-
ated vapor at 20° F
Enthalpy of R-12 satur-
ated liquid at 100° F

Refrigerating effect
Theoretical refrigerating
capacity of com-
pressor
Theoretical refriger-
ating capacity in
tons

= 16.52 cu ft/min

= 0.8921 lb/cu ft
= 16.52 x 0.8921
= 14.74 lb/min

= 80.49 Btu/lb

= 31.16 Btu/lb
= 49.33 Btu/lb

= 14.74 x 49.33
= 727.12 Btu/min
727.12

200
= 3.63 tons

Since specific volume is the reciprocal of
density, an alternate method of determining
the weight of refrigerant circulated per minute
by the compressor is to divide the piston
displacement of the compressor by the specific
volume of the suction vapor at the compressor
inlet.

When the volume of vapor to be circulated
per minute per ton for any given operating
conditions is known, the capacity of the com-
pressor in tons for the operating conditions in
question may be found by dividing the piston
displacement of the compressor by the volume
of vapor to be compressed per minute per ton.

Example 12-3. For the conditions of
Example 12-2, find (a) the weight of refrigerant
circulated per minute per ton ; (6) the volume of
vapor to be compressed per minute per ton;
and (c) the theoretical refrigerating capacity of
the compressor in tons.

Solution

(a) From Example 12-2,

refrigerating effect = 49.33 Btu/lb

Weight of refrigerant _ 200

circulated per minute ~~ 4933
Per ton = 4.05 lb/min

(b) From Table 16-3,
specific volume of
R-12 saturated vapor

at20°F = 1.121 cu ft/lb

Volume of vapor to be

compressed per =4.05 x 1.121

minute per ton = 4.55 cu ft/min

Theoretical refriger-
ating capacity of
compressor in tons

= 16.52 cu ft/min
_ 16.52
~ 4.55
= 3.63 tons

12-5. Actual Refrigerating Capacity. The

actual refrigerating capacity of a compressor is
always less than its theoretical capacity as
calculated in the previous examples. In the
preceding examples it has been assumed: (1)
that at each downstroke of the piston the cylinder
of the compressor fills completely with suction
vapor from the suction line and (2) that the
density of the vapor filling the cylinder is the
same as that in the suction line.

If these assumptions were correct, the actual
refrigerating capacity would be exactly equal to
the theoretical capacity. Unfortunately, this is
not the case. Because of the compressibility
of the refrigerant vapor and the mechanical
clearance between the piston and the valve plate
of the compressor, the volume of suction vapor
filling the cylinder during the suction stroke is
always less than the cylinder volume swept
through by the piston. Too, it will be shown
later that the density of the vapor filling the
cylinder is less than the density of the vapor
in the suction line. For these reasons, the actual
volume of suction vapor at suction line condi-
tions which is drawn into the cylinder of the
compressor is always less than the piston
displacement of the compressor and, therefore,
the actual refrigerating capacity of the com-
pressor is always less than its theoretical
capacity.

12-6. Total Volumetric Efficiency. The
actual volume of suction vapor compressed per
minute is the actual displacement of the com-
pressor. The ratio of the actual displacement of
the compressor to its piston displacement is
known as the total or real volumetric efficiency
of the compressor. Thus:

E -*

x 100

(12-2)

where E v = the total volumetric efficiency

V a = actual volume of suction vapor

compressed per minute
V v — the piston displacement of (he

compressor

PERFORMANCE OF RECIPROCATING COMPRESSORS 207

E v =

Actual weight of suction vapor

compressed x 100

Theoretical weight of suction vapor

compressed

When the volumetric efficiency of the com-
pressor is known, the actual displacement and
refrigerating capacity can be found as follows:

Ev (12-3)

V n - K. x

100

and

Actual

Theoretical

refrigerating = refrigerating x — ^- (12-4)
capacity capacity

Example 12-4. If the volumetric efficiency
of the compressor in Example 12-3 is 76%,
determine: (a) the actual volumetric displace-
ment (b) the actual refrigerating capacity.

Solution
(«) From Example 12-1,

piston displacement

Actual volumetric
displacement
(b) From Example 12-3,

theoretical refrigerating

capacity

Actual refrigerating
capacity

16.52 cu ft/min
16.52 x 0.76
12.66 cu ft/min

3.63 tons
3.63 x 0.76
2.76 tons

The actual refrigerating capacity of the
compressor ' may also be determined as in
Examples 12-2 and 12-3, if actual displacement
is substituted for piston displacement.
12-7. Factors Influencing Total Volumetric
Efficiency. The factors which tend to limit the
volume of suction vapor compressed per work-
ing stroke, thereby determining the volumetric
efficiency of the compressor, are the following:

1. Compressor clearance

2. Wiredrawing

3. Cylinder heating

4. Valve and piston leakage

12-8. The Effect of Clearance on Volumetric
Efficiency. Because of compressor clearance
and the compressibility of the refrigerant vapor,
the volume of suction vapor flowing into the
cylinder is less than the volume swept through
by the piston. As previously shown, at the end
of each compression stroke a certain amount of
vapor remains in the cylinder in the clearance

space after the discharge valves close. The
vapor left in the clearance space has been
compressed to the discharge pressure and, at
the beginning of the suction stroke, this vapor
must be re-expanded to the suction pressure
before the suction valves can open and allow
vapor from the suction line to flow into the
cylinder. The piston will have completed a part
of its suction stroke and the cylinder will already
be partially filled with the re-expanded clearance
vapor before the suction valves can open and
admit suction vapor to the cylinder. Hence,
suction vapor from the suction line will fill only
that part of the cylinder volume which is not
already filled with the re-expanded clearance
vapor.

In Fig. 12-3, V c is the total volume of the
cylinder when the piston is at the bottom of its
stroke. V a , which represents the clearance
volume, is the volume occupied by the clearance
vapor at the end of the compression stroke.
The difference between V c and V a then is the
volume of the cylinder swept through by the
piston each stroke. On the down stroke of the
piston, the clearance vapor expands from V a to
V b before the suction valves open. Therefore,
the part of the cylinder volume which is filled
with suction vapor during the balance of the
suction stroke is the difference between Vj,
and V e .

12-9. Theoretical Volumetric Efficiency.
The volumetric efficiency of a compressor due
to the clearance factor alone is known as the
theoretical volumetric efficiency. It can be
shown mathematically that the theoretical
volumetric efficiency varies with the amount of
clearance and with the suction and discharge
pressures. The reason for this is easily explained,
12-10. Effect of Increasing the Clearance.
If the clearance volume of the compressor is
increased in respect to the piston displacement,
the percentage of high-pressure vapor remaining
in the cylinder at the end of the compression
stroke will be increased. When re-expansion
takes place during the suction stroke, a greater
percentage of the total cylinder volume will be
filled with the re-expanded clearance vapor and
the volume of suction vapor taken in per stroke
will be less than when the clearance volume is
smaller. To obtain maximum volumetric
efficiency, the clearance volume of a vapor
compressor should be kept as small as possible.

208 PRINCIPLES OF REFRIGERATION

It should be noted that this does not hold
true for a reciprocating liquid pump. Since a
liquid is not compressible, the liquid left in the
clearance space at the end of the discharge
stroke has the same specific volume as the liquid
at the suction inlet. Therefore, there is no
re-expansion of the liquid in the clearance during
the suction stroke and the volume of liquid
taken in each stroke is always equal to the
volume swept by the piston, regardless of
clearance.

12-1 1. Variation with Suction and Discharge
Pressures. Increasing the discharge pressure or
lowering the suction pressure will have the same
effect on volumetric efficiency as increasing the
clearance. If the discharge pressure is increased,
the vapor in the clearance will be compressed
to a higher pressure and a greater amount of
re-expansion will be required to expand it to
the suction pressure. Likewise, if the suction
pressure is lowered, the clearance vapor must
experience a greater re-expansion in expanding
to the lower pressure before the suction valves
will open.

On the other hand, for a constant discharge
pressure, the amount of re-expansion that the
clearance vapor experiences before the suction
valves open diminishes as the suction pressure
rises. It is evident, then, that the volumetric
efficiency of the compressor increases as the
suction pressure increases and decreases as the
discharge pressure increases.
12-12. Compression Ratio. The ratio of the
absolute suction pressure to the absolute
discharge pressure is called the compression
ratio. Thus,

Absolute discharge pressure

JK — — r-: : . (12-3)

Absolute suction pressure
where R = the compression ratio.

Example 12-5. Calculate the compression
ratio of a R-12 compressor when the suction
temperature is 20° F and the condensing tem-
perature is 100° F.

Solution. From Table 16-3,
absolute pressure of R-12 satur-
ated vapor at 20° F = 35.75 psi

Absolute pressure of R-12
saturated vapor at 100° F = 131.6 psi

Compression ratio _ 131.6

- 35?75
= 3.69

Examination of Equation 12-3 indicates that
the compression ratio is increased by either
increasing the discharge pressure or lowering
the suction pressure, or both.

In the preceding section it was shown that
increasing the discharge pressure or lowering
the suction pressure decreases the volumetric
efficiency. It follows, then, that when the suction
and discharge pressures are varied in such a
direction that the compression ratio is increased,
the volumetric efficiency of the compressor
decreases. Likewise, decreasing the compression
ratio will increase the volumetric efficiency. For
a compressor of any given clearance, the volu-
metric efficiency varies inversely with the
compression ratio.

12-13. The Effects of Wiredrawing. Wire-
drawing is defined as a "restriction of area for a
flowing fluid, causing a loss in pressure by
(internal and external) friction without the loss
of heat or performance of work; throttling."*

In order to have a flow of vapor from the
suction line through the suction valves into the
compressor cylinder, there must be a pressure
differential across the valves sufficient to over-
come the spring tension of the valves and valve
weight and inertia. This means that the
suction vapor experiences a mild, throttling
expansion or drop in pressure as it flows
through the suction valves and passages of the
compressor. Therefore, the pressure of the
suction vapor filling the cylinder of the com-
pressor is always less than the pressure of the
vapor in the suction line. As a result of the
expanded condition of the vapor filling the
cylinder, the volume of suction vapor taken in
from the suction line each stroke is less than
if the vapor filling the cylinder was at the
suction line pressure.

A similar pressure differential is required
across the discharge valves in order to cause the
discharge vapor to flow through the valves
into the condenser. To provide the necessary
pressure differential across the discharge valves,
the vapor in the cylinder must be compressed to
a pressure somewhat higher than die actual
condensing pressure. The vapor left in the
clearance space at the end of the di charge
stroke will be at this higher pressure. To re-
expand from this higher pressure during the

* Asre Data Book, 1957-58 (page 39-27).

PERFORMANCE OF RECIPROCATING COMPRESSORS 209

suction stroke, the clearance vapor must suffer
a greater amount of re-expansion than if it had
been compressed only to the condensing pres-
sure. As a result of the greater expansion of
the clearance vapor, a larger portion of the
cylinder volume is filled with the re-expanded
clearance vapor during the down stroke of the
piston and the amount of suction vapor drawn
in from the suction line is reduced.

Unlike the other factors which determine
volumetric efficiency, wiredrawing is not directly
affected by the compression ratio. In general,
wiredrawing is a function of the velocity of the
refrigerant vapor flowing through the valves and
passages of the compressor. As the velocity of
the vapor through the valves is increased, the
effect of wiredrawing increases.

The refrigerant velocity through the valves of
a compressor depends upon the design of the
valves, the refrigerant used, and the speed of the
compressor.

Wiredrawing is greatest for those refrigerants
having the greatest specific volumes and the
lowest latent heat values because the volume
of vapor circulated per ton of refrigerating
capacity is greater. This accounts for the large
wiredrawing effect associated with R-12.

Increasing the speed of the compressor
increases the piston displacement. Hence, the
velocity of the vapor through the valves and the
effects of wiredrawing are increased as the rpm
are increased.

12-14. The Effects of Cylinder Heating.
Another factor which tends to reduce the
volumetric efficiency of the compressor is the
heating of the suction vapor in the compressor
cylinder. The suction vapor entering the
compressor cylinder is heated by heat con-
ducted from the hot cylinder walls and by
friction which results from the turbulence of the
vapor in the cylinder and from the fact that the
refrigerant vapor is not a perfect gas. The
heating causes the vapor to expand after
entering the cylinder so mat a smaller weight of
vapor will fill the cylinder and thereby still
further reduce the volume of vapor taken in
from the suction line.

Cylinder heating increases as the compression
ratio increases. At high compression ratios,
the work of compression is greater and the
discharge temperature is higher. This causes
a rise in the temperature of the cylinder walls

and other compressor parts so that the transfer
of heat to the suction vapor occurs at a higher
rate.

12-15. The Effect of Piston and Valve Leak-
age. Any back leakage of gas through either the
suction or discharge valves or around the piston
will decrease the volume of vapor pumped by
the compressor. Because of precision manu-
facturing processes, there is very little leakage of
gas around the pistons of a compressor in good
condition. However, since it is not possible to
design valves that will close instantaneously,
there is always a certain amount of back
leakage of gas through the suction and discharge
valves.

As the pressure in the cylinder is lowered at
the beginning of the suction stroke, a small
amount of high-pressure vapor in the head of
the compressor will leak back into the cylinder
before the discharge valves can close tightly.
Similarly, at the start of the compression stroke,
some of the vapor in the cylinder will flow
back through the suction valves into the, suction
line before the suction valves can close.

To assure prompt closing of the valves, both
the suction and discharge valves are usually
constructed of lightweight materials and are
slightly spring loaded. However, since the spring
tension increases wiredrawing, the amount of
spring loading is critical.

For any given compressor, the amount of
backleakage through the valves is a function of
the compression ratio and the speed of the
compressor. The higher the compression ratio,
the greater is the amount of valve leakage.
The effect of compressor speed on valve
leakage is discussed later.
12-16. Determining the Total Volumetric
Efficiency. The combined effects of all of the
foregoing factors on the volumetric efficiency of
the compressor varies with the design of the
compressor and with the refrigerant used.
Furthermore, for any one compressor the
volumetric efficiency is not a constant amount;
it changes with the operating conditions of the
system. Therefore, the total volumetric effi-
ciency of a compressor is difficult to predict
mathematically and can be determined with
accuracy only by actual testing of the compressor
in a laboratory.

However, the results of such tests indicate
that the volumetric efficiency of any one

210 PRINCIPLES OF REFRIGERATION

Volumetric efficiency

2 3 4 5 6 7 8 9 10 11 12 13 14
Compression ratio

Fig. 12-4. Effect of compression ratio on volumetric
efficiency of R-12 compressor.

compressor is primarily a function of the
compression ratio and, for any given com-
pression ratio, remains practically constant,
regardless of the operating range. It has been
determined also that compressors having the
same design characteristics will have approxi-
mately the same volumetric efficiencies, regard-
less of the size of the compressor.

The relationship between the compression
ratio and the volumetric efficiency of a typical
R-12 compressor is illustrated by the curve in
Fig. 12-4. In addition, in order to facilitate
future calculations, the average volumetric
efficiencies of a group of typical R-12 com-
pressors at various compression ratios are
given in Table 12-1. The values given are for
compressors ranging in size from 5 to 25 hp.
Smaller compressors will have slightly lower
efficiencies, whereas larger compressors will
have slightly higher efficiencies.
12-17. Variation in Compressor Capacity
with Suction Temperature. Compressor per-
formance and cycle efficiency will vary consider-
ably with the operating conditions of the system.
The most important factor governing the
capacity of the compressor is the vaporizing
temperature of the liquid in the evaporator,
that is, the suction temperature. The large
variations in compressor capacity which accom-
pany changes in the operating suction tempera-
ture are primarily a result of a difference in the
density of the suction vapor entering the
suction inlet of the compressor. The higher the
vaporizing temperature of the liquid in the
evaporator, the higher is the vaporizing pressure
and the greater is the density of the suction
vapor. Because of the difference in the density

of the suction vapor, each cubic foot of vapor
compressed by the compressor will represent a
greater weight of refrigerant when the suction
temperature is high than when the suction
temperature is low. This means that for any
given position displacement, the weight of
refrigerant circulated by the compressor per
unit of time increases as the suction temperature
increases.

The effect of suction temperature on com-
pressor capacity is best illustrated by an actual
example.

Example 12-6. Assuming 100% efficiency,
if the liquid reaches the refrigerant control at
100° F in each case, determine the weight
of refrigerant circulated per minute and the
theoretical refrigerating capacity of the com-
pressor in Example 12-1 when operating at each
of the following suction temperatures: (a) 10° F
and (b) 40° F.

Solution
(a) From Table 16-3,
density of R-12 satur-

ated vapor at 10° F

0.7402 lb/cu ft

From Example 12-1,
piston displace-
ment

= 16.52 cu ft/min

Weight of refrigerant
circulated per
minute at 10° F
suction

= 16.52 x 0.7402
= 12.23 lb/min

From Table 16-3,
enthalpy of R-12
saturated vapor at
10° F

= 79.36 Btu/lb

Enthalpy of R-12
liquid at 100° F

= 31.16 Btu/lb

Refrigerating effect

= 48.20 Btu/lb

Theoretical refriger-
ating capacity of
compressor at 10° F
suction, Btu/min

= 12.23 x 48.20
= 589.49 Btu/min

Theoretical refrigerat-
ing capacity in tons

(b) From Table 16-3,
density of R-12 satur-
ated vapor at 40° F

589.49

200

= 2.95 tons

= 1.263 lb/cu ft

From Example 12-1,
piston displace-
ment

= 16.52 cu ft/min

PERFORMANCE OF RECIPROCATING COMPRESSORS 211

Weight of refrigerant
circulated per minute

= 16.52 x 1.263

at 40° F suction

= 20.86 lb/min

From Table 16-3,

enthalpy of R-12
saturated vapor at
40° F

= 82.71 Btu/lb

Enthalpy of R-12
liquid at 100° F

= 31.16 Btu/lb

Refrigerating effect

= 51.55 Btu/lb

Theoretical refriger-

ating capacity of
compressor at 40° F
suction, Btu/min

= 20.86 x 51.55
= 1075.33 Btu/min

Theoretical refriger-

1075.33

ating capacity in

200

tons

= 5.38 tons

In analyzing the results of Example 12-6 the
following observations are of interest:

1. Although the piston displacement of the
compressor is the same in each case, the weight
of refrigerant circulated per minute by the
compressor increases from 12.23 lb/min to
20.86 lb/min when the operating suction
temperature is raised from 10° F to 40° F. The
increase in the weight of refrigerant circulated
results entirely from the greater density of the
suction vapor entering the suction inlet of the
compressor. In this instance, the percentage
increase in the weight of refrigerant circulated is

20.86 - 12.23

12.23

x 100 =70.5%

2. The theoretical refrigerating capacity of
the compressor at the 10° F suction temperature
is 2.95 tons, whereas at the 40° F suction
temperature, the capacity increases to 5.38 tons.
This represents an increase in refrigerating
capacity of

5.38 - 2.95
2^5 x 100 = 82.3%

Although the increased density of the suction
vapor at the higher suction temperature
accounts for the greater part of the increase in
compressor capacity, it is not the only reason
for it. As indicated, the increase in the weight
of refrigerant circulated is only 70.5%, whereas
the total increase in compressor capacity is
82.3%. The additional 1 1.8 %gain in capacity is
brought about by an increase in the refrigerating

effect of each pound of refrigerant circulated.
Although the actual gain in refrigerating effect
per pound is only 6.95 %, when this increase is
applied to the entire weight of refrigerant
circulated at the higher suction temperature, the
net gain in capacity over the original capacity
which can be attributed to the greater refriger-
ating effect is 11.8% (1.705 x 0.0695 = 1.823
and 1.283 - 1.705 = 0.118 or 11.8%).

The actual variation in compressor capacity
with changes in suction temperature is more
pronounced than that indicated by theoretical
computations. That is, the change in the actual
compressor capacity with variations in suction
temperature is always greater than the change
in the theoretical capacity. The reason for this
is that the compression ratio changes as the
suction temperature changes. When the vapor-
izing temperature increases while the condensing
temperature remains constant, the compression
ratio is decreased and the volumetric efficiency
of the compressor is improved. Hence, at the
higher suction temperature, in addition to
pumping a greater weight of refrigerant per unit
of volume, the volume of vapor pumped by the
compressor is also larger because of the im-
proved efficiency.

Example 12-7. Assuming that the satur-
ated discharge temperature is 100° F, determine
the actual refrigerating capacity of the com-
pressor in Example 12-6 when operating at each
of the suction temperatures in question.

Solution

(a) From Table 16-3,

absolute pressure corre-

sponding to 100° F

saturation temperature

131.6 psi

Absolute pressure corre-

sponding to 10° F

saturation temperature

= 29.35 psi

Compression ratio

131.6

29.35

= 4.47

From Table 12-1,

volumetric efficiency

= 76.3%

From Example 12-6,

theoretical refrigerating

capacity at 10° F

suction

= 2.95 tons

Actual refrigerating

capacity at 10° F

= 2.95 x 0.763

suction

= 2.22 tons

212 PRINCIPLES OF REFRIGERATION

(b) From Table 16-3,
absolute pressure corre-
sponding to 100° F satu-
ration temperature
Absolute pressure
corresponding to 40° F

= 131.6 psi

saturation temperature

= 51.68 psi

Compression ratio

131.6

31.68

= 2.55

From Table 12-1,

volumetric efficiency

= 85.7%

From Example 12-6,
theoretical refrigerating
capacity at 40° F
suction

= 5.38 tons

Actual refrigerating
capacity at 40° F
suction

= 5.38 x 0.857
= 4.61 tons

Whereas the theoretical increase in com-
pressor capacity is only 82.3%, the actual
increase in refrigerating capacity is

4.61 -2.22

2.22
= 107.7%

x 100

12-18. Effect of Condensing Temperature
on Compressor Capacity. In general, the
refrigerating capacity of the compressor de-
creases as the condensing temperature increases
and increases as the condensing temperature
decreases. The effect that the condensing
temperature has on compressor efficiency and
capacity can be evaluated by comparing the
results of the following example with those of
Examples 12-6 and 12-7.

Example 12-8. Determine the theoretical
and actual refrigerating capacities of the com-
pressor in Example 12-1 for each of the two
vaporizing temperatures given in Examples
12-6 and 12-7, if the condensing temperature in
each case is 120° F rather than 100° F.

Solution
(a) For the 10° F vaporizing temperature.
From Example 12-1,
piston displacement
of compressor

From Table 16-3, den-
sity of R-12 saturated
vapor at 10° F = 0.7402 lb/cu ft

= 16.52 cu ft/min

Theoretical weight of
refrigerant circulated
per minute by corn-
compressor

Refrigerating effect per
pound at 10° F
vaporizing and 120° F
condensing

Theoretical refriger-
ating capacity of
compressor

Theoretical refriger-

= 16.52 x 0.7402
= 12.23 lb

= 43.20 Btu/lb

= 12.23 x 43.20
= 527.34 Btu/min

527.34

ating capacity in tons

200

= 2.64 tons

From Table 16-3,

absolute suction

pressure

= 29.35 psi

Absolute discharge

pressure

= 171.8 psi

Compression ratio

171.8

29.35
= 5.85

= 66.5%

= 2.645 x
= 1.76

From Table 12-1,
volumetric efficiency

Actual refrigerating
capacity in tons

(ft) For the 40° F suction temperature.
From Example 12-1,
piston displacement
of compressor

From Table 16-3, den-
sity of R-12 saturated
vapor at 40° F

Theoretical weight of
refrigerant circulated
per minute by
compressor

0.665

= 16.52 cu ft/min

= 1.263 lb/cu ft

= 16.52 x 1.263
= 20.86 lb

Refrigerating effect per
pound at 40° F
evaporating and
condensing 120° F

Theoretical refriger-
ating capacity of
compressor

Theoretical refriger-
ating capacity in tons

From Table 16-3,
absolute suction
pressure

Absolute discharge
pressure

= 46.55 Btu/lb

= 20.86 x 46.55
= 971 Btu/min

971
~200
= 4.85

= 51.68 psi
= 171.8 psi

PERFORMANCE OF RECIPROCATING COMPRESSORS 213

Compression ratio

From Table 12-1,
volumetric efficiency

Actual refrigerating
capacity in tons

171.8
51.68
3.32

78.5%

4.85 x 0.785
3.81

Examining first the 10° F cycle, notice that
raising the condensing temperature from 100° F
to 120° F reduces the theoretical refrigerating
capacity of the compressor from 2.95 tons to
2.64 tons and the actual capacity from 2.22 tons
to 1.76 tons.

Since a 100% efficient compressor is assumed
to displace a theoretical volume of vapor equal
to its piston displacement and since the density
of the suction vapor entering the compressor
for any one vaporizing temperature is always
the same regardless of the condensing tempera-
ture, the theoretical weight of refrigerant
displaced by the compressor is the same at all
condensing temperatures, and therefore the
theoretical refrigerating capacity of the com-
pressor for any condensing temperature depends
only upon the refrigerating effect per pound of
refrigerant circulated. Hence, the difference
in the theoretical refrigerating capacity of the
compressor at the two condensing temperatures
results entirely from the difference in the
refrigerating effect per pound.

The reduction in actual compressor capacity
may be attributed to several factors: (1) a
reduction in the refrigerating effect per pound
and (2) a reduction in the volumetric efficiency
of the compressor.

Increasing the condensing temperature while
the suction temperature remains constant
increases the compression ratio and reduces
the volumetric efficiency of the compressor so
that the actual volume of vapor displaced by the
compressor per unit of time decreases. There-
fore, even though the density of the vapor
entering the compressor remains the same at
all condensing temperatures, the actual weight
of refrigerant circulated by the compressor per
unit of time decreases because of the reduction
in the quantity of vapor handled.

Increasing the condensing temperature in-
creases the isentropic discharge temperature. In
this instance, it is interesting to note (Fig. 7-7)
that the increase in the isentropic discharge

temperature is somewhat greater than that in the
condensing temperature. Whereas the increase
in condensing temperature is only 20° F
(120° - 100°), the increase in the discharge
temperature is 23.5° F (137.5° - 114°). This is
accounted for by the greater work of com-
pression at the higher compression ratio. Had
the condensing temperature been increased in
such a way that the compression ratio does not
change (by increasing the suction temperature
in proportion), the increase in the discharge
temperature would have been approximately
the same as that in the condensing tempera-
ture.

High discharge temperatures are undesirable
and are to be avoided whenever possible. The
higher the discharge temperature, the higher is
the average temperature of the cylinder walls
and the greater is the superheating of the
suction vapor in the compressor cylinder. In
addition to its adverse effect of compressor
efficiency, high discharge temperatures tend to
increase the rate of acid formation in the
system, cause carbonization of the oil in the
head of the compressor, and produce other
effects detrimental to the equipment.

The loss of compressor efficiency and capacity
resulting from an increase in the condensing
temperature of the cycle is more serious when
the suction temperature of the cycle is low than
when the suction temperature is high. The
desirability of operating a refrigerating system
at the lowest practical condensing temperature
has already been pointed out. This is of par-
ticular importance when the suction temperature
of the cycle is low and the compressor is already
operating at a relatively low efficiency.

When the cycle is operating at a 40° F
vaporizing temperature, increasing the con-
densing temperature from 100° F to 120° F
reduces the theoretical capacity of the com-
pressor from 5.38 tons to 4.85 tons and the
actual compressor capacity from 4.61 tons to
3.81 tons. The loss in theoretical capacity is

5.38 - 4.85

5.38

x 100 = 10%

The loss in actual compressor capacity amounts
to

4.61 - 3.81

4 . 61 x 100 = 17.4%

214 PRINCIPLES OF REFRIGERATION

For the 10° F cycle, the loss in theoretical
compressor capacity is

2.95 - 2.64

2.95

x 100 = 10.5%

and the loss in actual compressor capacity is
2.55 - 1.76

2.55

x 100 = 31%

Note that the loss in theoretical capacity
brought about by increasing the condensing
temperature is approximately the same for
both suction temperatures, whereas the loss in
actual compressor capacity is much greater at
the lower suction temperature. To a great
extent, it is the loss in volumetric efficiency that
causes the marked decrease in the actual
capacity of the compressor at the higher
condensing temperature. The change in volu-
metric efficiency for a given change in condens-
ing temperature becomes greater as the suction
temperature of the cycle decreases. This
accounts for the greater effect that a change in
condensing temperature has on compressor
capacity when the suction temperature is low.
12-19. Compressor Horsepower. The theore-
tical horsepower required to drive the compres-
sor may be found by multiplying the actual
refrigerating capacity of the compressor in tons
by the theoretical horsepower required per ton
for the operating conditions in question.

Example 12-9. Find the theoretical horse-
power required to drive the compressor in
Example 12-4.

Solution. From Example
12-4, actual refrigerating
capacity in tons = 2.76 tons

From Fig. 7-9, theoretical
horsepower required per ton = 0.965 hp

The theoretical horsepower = 2.76 x 0.965
required to drive the com-
pressor = 2.66 hp

Notice that it is the actual refrigerating
capacity of the compressor, rather than the
theoretical refrigerating capacity, which must
be used in determining the theoretical horse-
power requirements of the compressor.

The theoretical horsepower as calculated in
the preceding example is only an indication of
the power which would be required by a 100%

efficient compressor operating on an ideal
compression cycle and does not represent the
actual total horsepower which must be delivered
to the shaft of the compressor. In actual
practice, there are certain losses in power which
accrue because of the mechanical friction in the
compressor and because of the deviation of an
actual compression cycle from the ideal com-
pression cycle. Naturally, additional power
must be supplied to the compressor to offset
these losses. Therefore, the actual power
required by a compressor will always be greater
than the theoretical computations indicate.
12-20. Variation in Compressor Horse-
power with Suction Temperature. Although
the horsepower per ton of refrigerating capacity
diminishes as the suction temperature rises, the
horsepower required by the compressor may
either increase or decrease, depending upon
whether the work done by the compressor
increases or decreases.

The total amount of work done by the
compressor per unit of time in compressing the
vapor and, hence, the power required to drive
the compressor, is the function of only two
factors : (1) the work of compression per pound
of vapor compressed and (2) the weight of vapor
compressed per unit of time.

The amount of work which is done in
compressing the vapor from the suction
pressure to the discharge pressure varies with
the compression ratio. The greater the com-
pression ratio, the greater is the work of
compression. Therefore, when the suction
temperature is raised while the condensing
temperature remains the same, the compression
ratio becomes smaller and the work of com-
pression per pound is reduced. However, at the
same time, because of the greater density of the
suction vapor, the weight of vapor compressed
by the compressor per unit of time increases.
Since the saving in work done resulting from
the reduction in the work per pound is seldom
sufficient to outweigh the increase in the work
of the compressor because of the increase in the
weight of vapor compressed, raising the suction
temperature will usually increase the power
requirements of the compressor.

Example 12-10. Compute the theoretical
horsepower required by the compressor in
Example 12-7 at each of the suction tempera-
tures listed.

PERFORMANCE OF RECIPROCATING COMPRESSORS 215

Solution
(a) From Example 12-7,
actual refrigerating
capacity in tons at 10° F
suction temperature
From Fig. 7-9, theoreti-
cal horsepower per ton
at 10° suction and
100° F condensing
Theoretical horsepower
of compressor at 10° F
suction
(6) From Example 12-7,
actual refrigerating
capacity in tons at 40° F
suction temperature
From Fig. 7-9, theoreti-
cal horsepower per ton
at 40° F suction and
100° F condensing
Theoretical horsepower
of compressor at 40° F
suction

= 2.22 tons

= 1.13 hp

= 2.22 x 1.13
= 2.51 hp

= 4.61 tons

= 0.683 hp

= 4.61 x 0.683
= 3.15 hp

Although the horsepower per ton decreases
39.5 % as the suction temperature is raised from
10° F to 40° F, because of the increase in the
refrigerating capacity of the compressor, the
horsepower required by the compressor in-
creases from 2.51 hp to 3.15 hp. This represents
an increase in the power required of

3.15 -2.51

— 2 1T -X100=21%

The increase in compressor horsepower with
the suction temperature is relatively small in

Fig. 12-5. Curves illustrate the
effects of suction temperature
on the capacity and horse-
power of reciprocating com-
pressors.

comparison to the increase in compressor
capacity. In this instance, for a 30° F rise in
suction temperature, the capacity of the
compressor increased 107%, whereas com-
pressor horsepower increased only 21 %. The
average increase in compressor capacity per
degree of rise in suction temperature is 107%/
30° F or 3.21 %, whereas the increase in horse-
power amounts to only 0.7 % per degree of rise.

The relationship between compressor capacity
and the horsepower of the compressor at
various suction temperatures is shown by the
curves in Fig. 12-5. The curves are for a typical
R-12 compressor operating at a constant con-
densing temperature of 100° F.

As shown by the curve in Fig. 12-5 the
horsepower required by a R-12 compressor
increases as the suction temperature increases
up to a certain point at which the horsepower
required by the compressor is at a maximum.
On reaching this point, if the suction temperature
is further increased, the horsepower required
by the compressor diminishes. This is not true,
however, for compressors using ammonia as a
refrigerant. For compressors using ammonia,
the horsepower does not reach a maximum
value, but continues to increase indefinitely as
the suction temperature increases.

The suction temperature at which the horse-
power required by a R-12 compressors reaches
a maximum depends upon the condensing
temperature and increases as the condensing
temperature increases.

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Suction temperature

216 PRINCIPLES OF REFRIGERATION

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Condensing temperature

12-21. The Effect of Condensing Tempera-
ture on Compressor Horsepower. The

curves in Fig. 12-6 illustrate the relationship
between the horsepower required per ton of
refrigerating capacity, the actual refrigerating
capacity of the compressor, and the horsepower
required by the compressor at various condensing
temperatures when the suction temperature is
keptconstant. Note that, although the theoretical
horsepower required per ton increases as the
condensing temperature increases, the theoretical
horsepower required by any one compressor will
not increase in the same proportion. This is true
because the decrease in the refrigerating capa-
city of the compressor which is coincident with
an increase in the condensing temperature will
offset to some extent the increase in the horse-
power per ton.

For instance, according to Fig. 7-9, for a cycle
operating at a 10° F vaporizing temperature,
the theoretical horsepower required per ton
increases from 1 . 1 3 to 1 .52 when the condensing
temperature of the cycle is increased from 100° F
to 120° F. At the same time, Example 12-10
illustrates that the actual refrigerating capacity
of one particular compressor drops from 2.22
tons to 1.76 tons when the condensing tempera-
ture is raised from 100° F to 120° F. The
theoretical horsepower required by the com-
pressor at the 100° F condensing temperature is

1.13 x 2.22 = 2.51 hp

For the 120° F condensing temperature, the
theoretical horsepower required by the com-
pressor is

1.52 x 1.76 =2.68hp

4.6

4.5

4-4"

4.3 -S -

4.2 |
a.
E
o

4.1 <->

4.0
120*

2.6

2.2 I

i.8 i

o
1.4 1

m
1.0

Fig. 12-6. Curves illustrate
the effects of condensing tem-
perature on capacity and
horsepower of reciprocating
compressors.

12-22. Brake Horsepower. The total horse-
power which must be supplied to the shaft of the
compressor is called the brake horsepower and
may be computed from the theoretical horse-
power by application of a factor called over-all
compressor efficiency. The over-all efficiency is
an expression of the relationship of the theoreti-
cal horsepower to the brake horsepower in
percent. Written as an equation, the relationship
is

Thp

E °=w P xl0 °

and

Bhp =

Thp
E o l\00

(12-6)

(12-7)

where E = the over-all efficiency in percent
Thp = the theoretical horsepower
Bhp = the brake horsepower

Example 12-11. Determine the brake
horsepower required by the compressor in
Example 12-14, if the over-all efficiency of the
compressor is 80%.

Solution. From Example 12-4,

Thp

= 3.12 hp

Applying Equation 12-7, the

Thp

Bhp

3.12

086

= 3.9 hp

The over-all efficiency is sometimes broken
down into two components : (1) the compression
efficiency and (2) die mechanical efficiency. In
such cases, the relationship is

(12-8)

PERFORMANCE OF RECIPROCATING COMPRESSORS 217

where E e = the compression efficiency in per-
cent
E m = the mechanical efficiency in percent

so that

Bhp =

Thp

E c X E„

(12-9)

The compression efficiency of a compressor is
a measure of the losses resulting from the devia-
tion of the actual compression cycle from the
ideal compression cycle, whereas the mechanical
efficiency of the compressor is a measure of the
losses resulting from the mechanical friction in
the compressor. The principal factors which
bring about the deviation of an actual com-
pression cycle from the ideal compression cycle
are: (1) wiredrawing, (2) the exchange of heat
between the vapor and the cylinder walls, and
(3) fluid friction due to the turbulence of the
vapor in the cylinder and to the fact that the
refrigerant vapor is not an ideal gas. Notice
that the factors which determine the compression
efficiency of the compressor are the same as those
which influence the volumetric efficiency. As a
matter of fact, for any one compressor, the
volumetric and compression efficiencies are
roughly the same and they vary with the com-
pression ratio in about the same proportions.
For this reason, the brake horsepower required
per ton of refrigerating capacity can be approxi-
mated with reasonable accuracy by dividing the
theoretical horsepower per ton by the volu-
metric efficiency of the compressor and then
adding about 10% to offset the power loss due
to the mechanical friction in the compressor.
Written as an equation,

Bhp =

M(h d -h e ) x 1.1
42.42 x E„

(12-10)

Since the relationship between the various
factors which influence the compression effi-
ciency are difficult to evaluate mathematically,
the compression efficiency of a compressor can
be determined accurately only by actual testing
of the compressor.

12-23. Indicated Horsepower. A device fre-
quently used to determine the compression
efficiency is the indicator diagram. An indicator
diagram is a pressure-volume diagram of the
actual compression cycle of the compressor
which is produced during the actual testing of
the compressor.

A theoretical indicator diagram for an ideal
compression cycle is shown in Fig. 12-7. It has
been illustrated previously that the area under
a process diagram on a pressure-volume chart
is a measure of the work of the process. In
Fig. 12-7, notice that the area dDCd represents
the work done by the piston in compressing the
vapor during the isentropic process CD, and
that the area aADba represents the work done
by the piston in discharging the vapor from the
cylinder during the constant pressure process
DA, whereas the area aABa represents the work
done back on the piston by the vapor during
the isentropic re-expansion (of the clearance
vapor) process AB. Since the work of process
AB is work given back to the piston by the
fluid, the net work input to the compression
cycle is the sum of the work of processes CD
and DA, less the work of process AB. Therefore,
the net work of the compression cycle is
represented by the area BADCB, the total area
enclosed by the cycle diagram.

The work of the compression cycle as deter-
mined from the indicator diagram is called the
indicated work and the horsepower computed
from the indicated work is called the indicated
horsepower.

Since the indicator diagram illustrated in
Fig. 12-7 is a theoretical indicator diagram of
an ideal compression cycle, the indicated work
is the work of an ideal compression cycle and
the indicated horsepower computed from the
indicated work would, of course, be exactly

Volume

Fig. 12-7. Theoretical indicator diagram for an ideal
compression cycle.

218 PRINCIPLES OF REFRIGERATION

l«J

_4l^

1 "'

1
1

The relationship of the indicated horsepower
to the theoretical horsepower is

V b

V c '
Volume

Fig. 12-8. Theoretical indicator diagram for an
actual compression cycle.

equal to the theoretical horsepower. However,
in actual practice, since the indicator diagram
reproduces the true paths of the various
processes which make up the actual compression
cycle, the indicated work of the diagram is an
accurate measure of the actual work of the
compression cycle and, therefore, the ^indicated
horsepower computed on the basis of the
indicated work is the actual horsepower
required to do the work of the actual com-
pression cycle.

Care should be taken not to confuse indi-
cated horsepower with brake horsepower.
Although the indicated horsepower includes
the power required to offset the losses resulting
from the deviation of an actual compression
cycle from the ideal cycle, it does not include
the power required to overcome the losses
resulting from the mechanical friction in the
compressor. In other words, the indicated
horsepower takes into account the compression
efficiency but not the mechanical efficiency.
Hence, brake horsepower differs from indicated
horsepower in that the brake horsepower
includes the power required to overcome the
mechanical friction in the compressor, whereas
the indicated horsepower does not. The
horsepower necessary to overcome the mechani-
cal friction in the compressor is sometimes
referred to as the friction horsepower (Fhp), so
that

(12-11)

Thp

(12-12)

Bhp = Ihp + Fhp

An indicator diagram of an actual com-
pression cycle is shown in Fig. 12-8. The area
ABCD, enclosed by the cycle diagram, is, of
course, a measure of the work of the cycle.
An ideal cycle, AB'CD', is drawn in for com-
parison. Pressures P 1 and P 2 represent the
pressure of the vapor entering and leaving the
compressor. The areas above line P 2 and
below line P x represent the increased work of
the cycle due to wiredrawing. Notice that at
the end of the suction and discharge strokes
(points C and A), the piston velocity diminishes
to zero and the pressure of the vapor tends to
return to P x and P 2 , respectively. The other
deviations from the ideal cycle represent the
losses resulting from the heating of the vapor
in the compressor cylinder. Line BC indicates
the approximate volume of the suction vapor at
the end of the suction stroke, whereas line BC
represents the approximate volume of this same
weight of vapor in the suction line. The
deviation of the actual compression process
from the isentropic can be seen by comparing
the actual compression path CD to the isen-
tropic path CD'.

The direction of the periodic heat transfer
between the vapor and the cylinder walls at
various times and points in the cycle is indi-
cated by the arrows. The arrows pointing in
denote heat transfer from the cylinder walls to
the vapor, whereas arrows pointing out indicate
heat transfer from the vapor to the cylinder
walls.

The temperature of the cylinder walls of the
compressor will fluctuate around some mean
value which is between the suction and dis-
charge temperatures of the vapor. During the
latter part of the re-expansion process, during
the period in which the vapor is being admitted
to the cylinder, and during the initial part of the
compression stroke, the cylinder wall tempera-
ture is greater than the vapor temperature and
heat passes from the cylinder walls to the vapor.
During the latter part of the compression stroke,
during the discharge period, and during the
early part of the suction stroke, the temperature

PERFORMANCE OF RECIPROCATING COMPRESSORS 219

of the vapor exceeds the cylinder wall tempera-
ture and heat passes from the vapor to the
cylinder walls.

12-24. Isothermal vs. Isentropic Compres-
sion. Reference to Fig. 12-9 will show that if
the compression process in the compressor was
isothermal rather than isentropic the net work
of the compression cycle would be reduced even
though the work of the compression process
itself is greater for isothermal compression than
for isentropic compression. The reduction in
the work of the cycle which would be realized
through isothermal compression is indicated by
the crosshatched area in Fig. 12-9.

Isothermal compression is not practical for a
refrigeration compressor since it would result
in the discharge of saturated liquid from the
compressor. Furthermore, if a cooling medium
were available at a temperature low enough to
cool the compressor sufficiently to produce
isothermal compression, the cooling medium
could be used directly as the refrigerant and
there would be no need for the refrigeration
cycle.

12-25. Water-Jacketing the Compressor
Cylinder. Any heat which is given up by the
compressor cylinder to some external cooling
medium represents, in effect, heat given up by
the vapor during the compression process.
Cooling of the vapor during compression causes
the path of the compression process to shift
from the isentropic path toward an isothermal
path. Of course, the greater the amount of
cooling, the greater will be the shift toward the
isothermal.

If the temperature of the air surrounding the
compressor were exactly the same as the
temperature of the compressor cylinder, there
would be no transfer of heat from the cylinder
to the air and any heat given up by the vapor
to the cylinder would be eventually reabsorbed
by the vapor and the compression process
would be approximately adiabatic. However,
since there is nearly always some transfer of
heat from the compressor to the surrounding
air, compression is usually polytropic rather
than isentropic. For an air-cooled compressor,
the transfer of heat to the air will be slight and,
therefore, the value of the polytropic com-
pression exponent, n, will very nearly approach
the isentropic compression exponent, k. Hence,
the assumption of isentropic compression for

the ideal cycle is ordinarily not too much in
error for an air-cooled compressor.

Water-jacketing of the compressor cylinder
results in lowering the temperature of the
cylinder walls, and cooling of the vapor during
compression will be greater for the compressor
having a water jacket. Too, cylinder heating is
reduced and the vapor is discharged from the
compressor at a lower temperature. All of this
has the effect of reducing the work of the
compression cycle. However, the gain is
usually not sufficient to warrant the use of a
water jacket on most compressors, particularly
compressors designed for R-12. For the most
part, water-jacketing of the compressor is
limited to compressors designed for use with
refrigerants which have unusually high dis-
charge temperatures, such as ammonia. Even
then, the purpose of the jacketing is not so
much for increasing compressor efficiency as
it is to reduce the rate of oil carbonization and
the formation of acids, both of which increase
rapidly as the discharge temperature increases.
12-26. Wet Compression. Wet compression
occurs when small particles of unvaporized
liquid are entrained in the suction vapor
entering the compressor. However, theoretical
computations indicate that wet compression
will bring about desirable gains in compression
efficiency and reduce the work of compression.
This would be true if the small particles of liquid
vaporized during the actual compression of the
vapor. However, in actual practice, this is not
the case. Since heat transfer is a function of
time and since compression of the vapor in a
modern high-speed compressor takes place very

Volume
Fig. 12-9. Isentropic vs. isothermal compression.

220

PRINCIPLES OF REFRIGERATION

rapidly, there is not sufficient time for the liquid
to completely vaporize during the compression
stroke. Hence, some of the liquid particles
remain in the vapor in the clearance volume
and vaporize during the early part of the
suction stroke. This action reduces the volu-
metric efficiency of the compressor without
benefit of the return of work to the piston by
the expansion of the vaporizing particles.

A result similar to this is encountered when
excessive cooling of the cylinder reduces the
temperature of the vapor in the clearance below
the saturation temperature corresponding to
the discharge. Some of the clearance vapor will
condense and the particles of liquid formed will
vaporize during the early part of the suction
stroke.

12-27. The Effect of Compressor Clearance
on Horsepower. Theoretically, the clearance
of the compressor has no effect on the horse-
power, since the work done by the piston in
compressing the clearance vapor is returned to
the piston as the clearance vapor re-expands
at the start of the suction stroke. However, since
the refrigerant vapor is not an ideal gas, there
is some loss of power in overcoming the internal
friction of the fluid so that the power returned
to the piston during the re-expansion of the
clearance vapor will always be less than
the power required to compress it. Hence, the
clearance does have some, although probably
slight, effect on the power requirements.
12-28. Compressor Speed. Since the speed of
rotation is one of the factors determining piston*
displacement (Equation 12-1), the capacity of
the compressor changes considerably when the
speed of the compressor is changed. If the
speed of the compressor is increased, the piston
displacement is increased and the compressor
displaces a greater volume of vapor per unit of
time. Theoretically, based on the. assumption
that the volumetric efficiency of the compressor
remains constant, the capacity of the compressor
varies in direct proportion to the speed change.
That is, if the speed of the compressor is
doubled, the piston displacement and capacity
of the compressor are also doubled. Likewise,
if the speed of the compressor is reduced, the
piston displacement and capacity of the
compressor are reduced in the same propor-
tion. However, the volumetric efficiency of the
compressor does not remain constant during

speed changes, and therefore the change in
compressor capacity will not be proportional
to the speed change.

The variation in the volumetric efficiency
with changes in the speed of rotation is brought
about principally by changes in the -effects of
wiredrawing, cylinder heating, and the back
leakage of gas through the suction and discharge
valves.

The amount of back leakage through the
valves in percent per cubic foot of vapor dis-
placed is at a maximum at low compressor
speeds and decreases as the speed of the com-
pressor is increased. Cylinder heating, too, is
greatest at low compressor speeds. On the
other hand, the effect of wiredrawing is at a
minimum at low speeds and increases as the
speed increases because of the increase in the
velocity of the vapor passing through the valves.
Hence, as the speed of rotation increases, the
volumetric efficiency of the compressor due to
the cylinder heating and valve leakage factors
increases, while, at the same time, the volumetric
efficiency due to the wiredrawing factor de-
creases. It follows, then, that there is one
critical speed of rotation at which the combined
effect of these factors are at a minimum and the
volumetric efficiency is at a maximum. From
an efficiency standpoint, this is the speed at
which the compressor should be operated. At
speeds higher than this critical speed, the
volumetric efficiency of the compressor diminish
because the loss of efficiency due to the wire-
drawing effect will be greater than the gain
resulting from the decrease in the effect of
cylinder heating and valve leakage. Likewise, at
speeds below the critical speed, the volumetric
efficiency will be lower because the losses
accruing from the increase in cylinder heating
and valve leakage will be greater than the gain
resulting from the decrease in the wiredrawing
losses.

The critical speed will vary with the design of
the compressor and with the refrigerant used,
and can best be determined by actual test of the
compressor.

It is general practice in the design of modern
high-speed compressors to use large valve ports
in order to reduce the wiredrawing effect to a
practical minimum. These large openings in the
valve-plate tend to increase the clearance
volume and decrease the volumetric efficiency

PERFORMANCE OF RECIPROCATING COMPRESSORS 221

due to the clearance factor, but the advantages
accruing from the reduction in the wiredrawing
effect more than offsets the loss of efficiency due
to the greater clearance. This is particularly
true where the power requirements are con-
cerned, since the loss of power due to wire-
drawing is much greater than the loss of power
due to the clearance factor.
12-29. Mechanical Efficiency. The mechanical
friction in the compressor varies with the speed
of rotation, but for any one speed, the mechan-
ical friction, and therefore the friction horse-
power, will remain practically the same at all
operating conditions. Since the friction horse-
power remains the same, it follows that the
mechanical efficiency of the compressor depends
entirely upon the loading of the compressor.
As the total brake horsepower of the compressor
increases due to loading of the compressor, the
friction horsepower, being constant, will become
a smaller and smaller percentage of the total
horsepower .and the mechanical efficiency will
increase. It is evident that the mechanical
efficiency of the compressor will be greatest
when the compressor is fully loaded. The
mechanical efficiency of the compressor will
vary with the design of the compressor and
with compressor speed. An average compressor
of good design operating fully loaded at a
standard speed should have a mechanical
efficiency somewhat above 90%.
12-30. The Effect of Suction Superheat on
Compressor Performance. It has been shown
that superheating of the suction vapor causes
the vapor to reach the compressor in an
expanded condition. Therefore, when the
vapor reaches the compressor in a superheated
condition, the weight of refrigerant circulated
by the compressor per minute is less than when
the vapor reaches the compressor saturated.
Whether or not the reduction in the weight of
refrigerant circulated by the compressor reduces
the refrigerating capacity of the compressor
depends upon whether or not the superheating
produces useful cooling. When the super-
heating produces useful cooling, the gain in
refrigerating capacity resulting from the increase
in the refrigerating effect per pound is usually
sufficient to offset the loss in refrigerating
capacity resulting from the reduction in the
weight of refrigerant circulated. On the other
hand, when the superheating produces no

useful cooling, there is no offsetting gain in
capacity and the refrigerating capacity of the
compressor is reduced in inverse proportion to
the increase in the specific volume of the
suction vapor at the compressor inlet.

Regardless of whether or not the superheating
produces useful cooling, the horsepower
required to drive any one compressor is
practically the same for a superheated cycle as
for a saturated cycle. It was shown in Section
8-4 that, when superheating of the vapor
produces useful cooling, both the horsepower
required per ton and the refrigerating capacity
of the compressor are the same for the super-
heated cycle as for the saturated cycle. It
follows, then, that the horsepower required by
any one compressor will be the same for both
cycles. On the other hand, when superheating
of the vapor produces no useful cooling, the
horsepower per ton is greater than for the
saturated cycle. However, at the same time,
the refrigerating capacity of the compressor is
less for the superheated cycle and the increase
in the horsepower required per ton is more or
less offset by the reduction in compressor
capacity, so that the horsepower required by the
compressor is still approximately the same as
for the saturated cycle. Notice that, although
the horsepower required by any one compressor
is not appreciably changed by the superheating
of the suction vapor, when the superheating
does not produce useful cooling, the refriger-
ating capacity and efficiency of the compressor
are materially reduced. This is particularly
true when the compressor is operating at a low
suction temperature. It should be noted also
that superheating of the suction vapor reduces
the amount of cylinder heating and the efficiency
of the compressor is increased to some extent.
12-31. The Effect of Subcooling on Com-
pressor Performance. When subcooling of
the liquid refrigerant is accomplished in such a
way that the heat given up by the liquid leaves
the system, the specific volume of the suction
vapor at the compressor inlet is unaffected by
the subcooling and the weight of refrigerant
circulated per minute by the compressor is the
same as when no subcooling takes place. Since
the refrigerating effect per pound is increased by
the subcooling, the capacity of the compressor is
increased by an amount equal to the amount of
subcooling. Notice that the increase in the

222 PRINCIPLES OF REFRIGERATION

refrigerating capacity of the compressor result-
ing from the subcooling is accomplished without
increasing the power requirements of the com-
pressor. Therefore, subcooling improves com-
pressor efficiency, provided the heat given up
during the subcooling leaves the system.

When the heat given up during the subcooling
does not leave the system, as when a heat
exchanger is used, the gain in capacity due to
the subcooling is approximately equal to the
loss in capacity due to the superheating, and
the refrigerating capacity of the compressor is
very little affected. However, there is some
small gain in compressor efficiency, since the
superheating of the vapor in the heat exchanger
reduces the effect of cylinder heating.
12-32. Compressor Rating and Selection.
As previously stated, mathematical evaluation
of all the factors which influence compressor
performance is not practical. Hence, com-
pressor capacity and horsepower requirements
are determined accurately only by actual
testing of the compressor. Table R-10A is a
typical compressor rating table supplied by the
compressor manufacturer for use in compressor
selection. The ratings have been determined by
actual testing of the compressor under operating
conditions set forth in the compressor testing
and rating standards of the American Society
of Heating, Refrigerating, and Air Conditioning
Engineers (see Table R-10B).

It has been shown in the foregoing sections
that both the refrigerating capacity and the
horsepower requirements of a compressor vary
with the condition of the refrigerant vapor
entering and leaving the compressor. Notice
in Table R-10A that compressor refrigerating
capacities (Btu/hr) and horsepower require-
ments are listed for various saturated suction
and discharge temperatures. The saturated
suction temperature is the saturation tempera-
ture corresponding to the pressure of the vapor
at the suction inlet of the compressor, and the
saturated discharge temperature is the saturation
temperature corresponding to the pressure of
the vapor at the discharge of the compressor.

Although compressor ratings are based on
the saturated suction and discharge tempera-
tures, ASHRAE test standards require a certain
amount of suction superheat and specify that
the actual temperature of the suction vapor
entering the compressor be those listed in

Table R-10C. Since the compressor ratings
given in Table R-10A are in accordance with
ASHRAE standards, it follows that in order to
realize the listed ratings, the suction vapor must
enter the suction inlet of the compressor at the
conditions shown in Table R-10C. For example,
for a compressor operating at a saturated
suction of —40° F, the suction vapor should
enter the compressor at a temperature of 35° F
(superheated 75° F from -40° F to 35° F), if
the listed rating is to be obtained. Likewise, for
a compressor operating at a saturated suction
of 40° F, the actual temperature of the suction
vapor entering the compressor should be 65° F
(superheated 25° F from 40° F to 65" F).
Where the actual temperature of the suction
vapor is less than that indicated in Table
R-10C, the tabulated rating is corrected by
using an appropriate multiplier to obtain the
actual compressor capacity. The multipliers
given in Table R-10D correct the ratings to a
basis of no superheat for the saturated condition
listed. Where the actual suction vapor tem-
perature is intermediate between saturation and
the temperature shown in Table R-10C, the
multiplier is corrected accordingly.

The superheating is assumed to occur in the
evaporator, in the suction line inside the re-
frigerated space, or in a liquid-suction heat
exchanger so that the superheat produces useful
cooling (Section 8-4). Superheating which
occurs outside the refrigerated space should
be disregarded with respect to the tabulated
ratings.

The superheating requirement of the
ASHRAE standards at first appears to compli-
cate unnecessarily the compressor rating and
selection procedure. However, this is not the
case. The superheating requirement is very
realistic in that the amount of superheating
specified in the rating standards very nearly
approaches that amount which would normally
be expected in a well-designed application.
Hence, the effect of the superheating require-
ment is to cause the compressor to be rated
under conditions similar to those under which
the compressor will be operating in the field.
For this reason, except in unusual cases, no
appreciable error will occur if the compressor
ratings given in Table R-10A are used without
correction of any kind. Furthermore, com-
pressor capacity requirements are not usually

PERFORMANCE OF RECIPROCATING COMPRESSORS 223

critical within certain limits. There are several
reasons for this. First of all, the methods of
determining the required compressor capacity
(cooling load calculations) are not in themselves
exact. Too, it is seldom possible to select a
compressor which has exactly the required
capacity at the design conditions. Another
reason that compressor capacity is not critical
within reasonable limits is that the operating
conditions of the system do not remain constant
at all times, but vary from time to time with the
loading of the system, the temperature of the
condensing medium, etc. General practice is to
select a compressor having a capacity equal to
or somewhat in excess of the required capacity
at the design operating conditions.

It was shown in Chapter 8 that subcooling of
the liquid increases the refrigerating effect per
pound and thereby increases compressor
capacity. With regard to subcooling, the
ratings given in Table R-10A are based on
saturated liquid approaching the refrigerant
control, that is, no subcooling. Where the
liquid is subcooled by external means (Section
8-7), the capacity of the compressor may be
increased approximately 2% for each 5°F of
subcooling. Here again, for reasons outlined in
the preceding paragraph, the effect of subcooling
is usually neglected in selecting the compressor.

To select a compressor for a given application,
the following data are needed:

1. The required refrigerating capacity
(Btu/hr)

2. The design saturated suction temperature

3. The design saturated discharge temperature

Naturally, the required refrigerating capacity
is the average hourly load as determined by the
cooling load calculations. However, if an
evaporator selection is made prior to the
compressor selection, the compressor should be
selected to match the evaporator capacity rather
than the calculated load. The reasons for this
are discussed in Chapter 13.

The design saturated suction temperature
depends upon the design conditions of the
application. Specifically, it depends upon the
evaporator temperature (the saturation tem-
perature of the refrigerant at the evaporator
outlet) and upon the pressure loss in the suction
line. For instance, assume an evaporator
temperature of 28° F and a suction line pressure

loss of approximately 3 psi. From Table 16-3,
the saturation pressure of Refrigerant- 12 corre-
sponding to a temperature of 28° F is 41 .59 psia.
Allowing for the 3 psi pressure loss in the
suction line, the pressure of the vapor at the
suction inlet of the compressor is 38.39 psia
(41.59 - 3). From Table 16-3, the saturation
temperature corresponding to a pressure of
38.59 psia, and, therefore, the saturated suction
temperature, is approximately 24° F.

The design saturated discharge temperature
depends primarily on the size of the condenser
selected and upon the quantity and temperature
of the available condensing medium. Methods of
condenser selection are discussed in Chapter 14.
12-33. Condensing Unit Rating and Selec-
tion. Since condensing unit capacity depends
upon the capacity of the compressor, methods
of rating and selecting condensing units are
practically the same as those for rating and
selecting compressors. The only difference is
that, whereas compressor capacities are based
on the saturated suction and discharge tempera-
tures, condensing unit capacities are based on
the saturated suction temperature and on the
quantity and temperature of the condensing
medium. Since the size of the condenser is
fixed at the time of manufacture for any given
condenser loading, the only variables deter-
mining the saturation temperature at the
discharge of the compressor (and therefore the
capacity of the compressor at any given suction
temperature) is the quantity and temperature of
the condensing medium. For air-cooled con-
densing units, when the quantity of the air
passing over the condenser is fixed by the fan
selection at the time of manufacture, the only
variable determining the capacity of the con-
densing unit, other than the suction temperature,
is the ambient air temperature (temperature of
the air entering the condenser). Hence, ratings
for air-cooled condensing units are based on
the saturated suction temperature and the
ambient air temperature.

Ratings for water-cooled condensing units
are based on the saturated suction temperature
and on the entering and leaving water tempera-
tures.* Typical capacity ratings for air-cooled

* For any given condenser loading and entering
water temperature, the leaving water temperature
depends only on the quantity of water (gallons per
minute) flowing through the condenser.

224 PRINCIPLES OF REFRIGERATION

and water-cooled condensing units are given in
Tables R-ll and R-12, respectively.

Example 12-12. A certain refrigeration
application has a calculated cooling load of
33,000 Btu/hr. If the design saturated suction
and discharge temperatures are 20° F and
100° F, respectively, select a compressor from
Table R-10A which will meet the requirements
of the application.

Solution. Locate the desired saturated dis-
charge temperature in the first column of the
table (100° F). Next, in the second column,
locate the desired saturated suction tempera-
ture and read to the right until a compressor
having a capacity equal to or somewhat in
excess of the desired capacity is found. Select
compressor, Model #5F20, which has a capacity
of 34,000 Btu/hr at 1450 rpm.

Example 12-13. Assume that the design
saturated suction in Example 12-12 is 0° F and
make a new compressor selection.

Solution. Using the procedure outlined in the
solution to Example 12-12, select compressor
Model #5F30 which has a capacity of 36,000
Btu/hr at 1750 rpm.

Example 12-14. Determine the capacity of
compressor Model #5F20 when operating at a
23° F saturated suction temperature, if the
saturated discharge temperature is 100° F.

Approximate capacity
of compressor at a 23° F
saturated suction

= 34,000 + 2780
= 37,380 Btu/hr

Solution. From Table R-10A,

Compressor capacity
at 30° F suction

= 44,200 Btu/hr

Compressor capacity
at 20° F suction

= 34,600 Btu/hr

Capacity change per
10° F change in saturated
suction temperature

Average capacity change
per ° F change in suction
temperature

= 44,200 - 34,600
= 9600 Btu/hr

9600
10

Total capacity change
for 3° F change in suction
temperature

= 960 Btu/hr

= 960 x 3
= 2780 Btu/hr

Example 12-15. A certain refrigeration
application has a calculated cooling load of
8750 Btu/hr and the ambient temperature is
90° F. If the design saturated suction tempera-
ture is 20° F, select an air-cooled condensing
unit which will satisfy the requirements of the
application.

Solution. From Table R-ll, select a 1 hp
condensing unit having a capacity of 9340
Btu/hr at the prescribed conditions.

PROBLEMS

1. A four-cylinder reciprocating compressor
having a 2 in. bore and a 2.5 in. stroke is
rotating at 100 rpm. Compute the piston
displacement in cubic feet per minute.

Arts. 18.18 cfm

2. Assume the compressor in Problem 12-1 is
operating on the cycle described in Problem 7-1
and compute the theoretical refrigerating
capacity of the compressor in Btu/hr.

Ans. 2.47 tons

3. Using the conditions of Problem 12-2,
determine:

(a) The volumetric efficiency of the com-
pressor.

(b) The actual refrigerating capacity of the
compressor in tons. Ans. 1.52 tons

(c) The brake horsepower required per ton
(allow 10% for mechanical friction and
assume the compression efficiency is the
same as the volumetric efficiency).

Ans. 2.7 hp/ton

(d) The total brake horsepower required to
drive the compressor. Ans. 4.1 hp

4. From the compressor rating tables, select a
compressor which will satisfy the following
conditions:

(a) Required capacity 24,900 Btu/hr

(b) Design saturated suction temperature
10° F

System Equilibrium
and Cycling
Controls

13-1. System Balance. In the designing of a
refrigerating system, one of the most important
considerations is that of establishing the proper
relationship or "balance" between the vapor-
izing and condensing sections of the system.
It is important to recognize that whenever an
evaporator and a condensing unit are connected
together in a common system, a condition
of equilibrium or "balance" is automatically
established between the two such that the rate
of vaporization is always equal to the rate of
condensation. That is, the rate at which the
vapor is removed from the evaporator and
condensed by the condensing unit is always
equal to the rate at which the vapor is produced
in the evaporator by the boiling action of the
liquid refrigerant. Since all the components in
a refrigerating system are connected together in
series, the refrigerant flow rate through all
components is the same. It follows, therefore,
that the capacity of all the components must be
of necessity the same.

Obviously, then, where the system com-
ponents are selected to have equal capacities at
the system design conditions, the point of
system equilibrium or balance will occur at the
system design conditions. On the other hand,
when the components selected do not have
equal capacities at the system design conditions,
system equilibrium will be established at
operating conditions other than the system

design conditions and the system will not
perform satisfactorily.

In any event, it is important to understand
that, regardless of the equipment selected, the
system will always establish equilibrium at some
conditions such that all the system components
will have equal capacity. Hence, whether or not
system equilibrium is established at the sys-
tem design conditions depends entirely upon
whether or not the equipment is selected to have
approximately equal capacities at the system
design conditions. This concept is best illu-
strated through the use of a series of examples.

Example 13-1. A walk-in cooler, having a
calculated cooling load of 11,000 Btu/hr, is to
be maintained at 35° F. The desired evaporator
TD is 12° F and the ambient temperature is
90° F. Allowing 3° F (equivalent to approxi-
mately 2 lb) for the pressure drop in the suction
line (see Section 12-32, select an air-cooled
condensing unit and a unit cooler from manu-
facturer's catalog data.

Solution. Since the design space temperature
is 35° F and the design evaporator TD is 12° F,
the design evaporator temperature is 23° F
(35° F - 12° F). When we allow for a 3° F
loss in the suction line resulting from pressure
drop, the saturation temperature at the com-
pressor suction is 20° F (23° F - 3° F).

From Table R-ll, select lj hp condensing
unit, which has a capacity of 12,630 Btu/hr at a
20° F saturated suction and 90° F ambient air
temperature. Although the condensing unit
capacity is somewhat in excess of the calculated
load of 11,000 Btu/hr, it is sufficiently close to
make the condensing unit acceptable. How-
ever, to assure proper system balance, the unit
cooler selection must now be based on the
condensing unit capacity of 12,630 Btu/hr
rather than on the calculated load of 11,000
Btu/hr. * Hence, a unit cooler having a capacity
of approximately 12,630 Btu/hr at a 12° F TD
is required.

From Table R-8, unit cooler Model #105 has
a capacity of 10,500 Btu/hr at a 10° F TD.
Using the procedure outlined in Section 11-23,
it can be determined that this unit cooler will
have a capacity of 1 2,600 Btu/hr when operating

* Either the evaporator or the condensing unit
may be selected first. However, once either one has
been selected, the other must be selected for approxi-
mately the same capacity.

225

226 PRINCIPLES OF REFRIGERATION

(Compressor; -

\*w

1 >

5° 10° 15° 20°
Suction temperature

J I I I I L

25°

32°

27° 22° XT 12°
Evaporator TD
Fig. 13-1. Graphic analysis of system balance.

at a 12° F TD. Since this is very close to the
condensing unit capacity, the unit cooler is
ideally suited for the application.
13-2. Graphical Analysis of System Equi-
librium. For any particular evaporator and
condensing unit connected together in a
common system, the relationship established
between the two, that is, the point of system
balance, can be evaluated graphically by
plotting evaporator capacity against condensing
unit capacity on a common graph. Using data
taken from the manufacturers' rating tables,
condensing unit capacity is plotted against
suction temperature, whereas evaporator capa-
city is plotted against evaporator TD. A
graphical analysis of the system described in
Example 13-1 is shown in Fig. 13-1.

In order to understand the graphical analysis
of system equilibrium in Fig. 1 3-1 , it is important
to recognize that, for any given space tempera-
ture, there is a fixed relationship between the
evaporator TD and the compressor suction
temperature. That is, for any given space
temperature, once the evaporator TD is
selected, there is only one possible suction
temperature which will satisfy the design
conditions of the system.

Notice that, in Example 13-1, for the design
space temperature of 35° F and assuming a
3° F loss in the suction line, the only possible
suction temperature that can coexist with the
design evaporator TD of 12° F is 20° F. In this

instance, the evaporator TD will be 12° F when,
and only when, the suction temperature is 20° F.
Any suction temperature other than 20° F will
result in an evaporator TD either greater or
smaller than 12° F. For example, assume a
suction temperature of 25° F. Adding 3° F to
allow for the suction line loss, the evaporator
temperature is found to be 28° F (25° F + 3° F).
Then subtracting the evaporator temperature
from the space temperature, it is determined
that the evaporator TD will be 7° F (35° F -
28° F) when the suction temperature is 25° F.
Using the same procedure, it can be shown that
if the suction temperature is reduced to 15° F,
the evaporator TD will increase to 17° F, and
when the suction temperature is 10° F, the
evaporator TD will be 22° F, and so on.

Apparently, then, raising or lowering the
suction temperature always brings about a
corresponding adjustment in the evaporator TD.
Provided that the space temperature is kept
constant, raising the suction temperature
reduces the evaporator TD, whereas lowering
the suction temperature increases the evaporator
TD.

With regard to Fig. 13-1, the following
procedure is used in making a graphical analysis
of the system equilibrium conditions:

1. On graph paper, lay out suitable scales
for capacity (Btu/hr), suction temperature (° F),
and evaporator TD (° F). The horizontal lines
are used to represent capacity, whereas the
vertical lines are given dual values, representing
both suction temperature and evaporator TD.
The latter is meaningful, however, only when
the suction temperature and evaporator TD
scales are so correlated that the two conditions
which identify any one vertical line are condi-
tions which, at the design space temperature,
actually represent conditions that will occur
simultaneously in the system. The procedure
for correlating the suction temperature and
evaporator TD scales was discussed in the
preceding paragraphs.

2. Using manufacturer's catalog data, plot
the capacity curve for the condensing unit.
Since condensing unit capacity is not exactly
proportional to suction temperature, the con-
densing unit capacity curve will ordinarily have
a slight curvature. Hence, for accuracy, a
capacity point is plotted for each of the suction

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 227

temperatures listed in the table and these points
are connected with the "best-fitting" curve.

3. From the evaporator manufacturer's cata-
log data, plot the evaporator capacity curve.
Since evaporator capacity is assumed to be
proportional to the evaporator TD, the evapo-
rator capacity curve isa straight line, the position
and direction of which is adequately established
by plotting the evaporator capacity at any two
selected TDs. The evaporator capacity at any
other TD will fall somewhere along a straight
line drawn through these two points. In Fig.
13-1, evaporator TDs of 7°F and 12° F are
used in plotting the two capacity points required
to establish the evaporator capacity curve.

Notice in Fig. 13-1, that as the suction
temperature increases, the evaporator TD
decreases. This means, in effect, that as the
suction temperature increases the capacity of
the evaporator decreases while the capacity of
the condensing unit (compressor) increases.
Likewise, as the suction temperature decreases,
the capacity of the evaporator increases while
the capacity of the condensing unit decreases.
The intersection of the two capacity curves
indicates the point of system equilibrium. In
this instance, because the evaporator and
condensing unit have been selected to have
equal capacities at the system design conditions,
the point of system equilibrium occurs at the
system design conditions (12° F TD and 20° F
suction temperature). Although the total
system capacity is somewhat greater than the
calculated load, the difference is not sufficiently
great to be of any particular consequence, and
means only that the system will operate fewer
hours out of each 24 than was originally
anticipated.* The relationship between system
capacity and the calculated load is discussed
more fully later in the chapter.

It has already been pointed out that where the
evaporator and condensing unit selected do not
have equal capacities at the system design
conditions, the point of system equilibrium will

* For simplicity, the heat given off by the evapo-
rator fan motor has been neglected. If this heat is
added to the cooling load, the total system capacity
would be almost exactly equal to the calculated load.
It is not often that equipment can be found which so
nearly meets the requirements of an application as in
this instance.

occur at conditions other than the design
conditions. For instance, assume that unit
cooler Model #UC-120, rather than Model
#UC-105, is selected in the foregoing example.
At the design evaporator TD of 12° F, this unit
cooler has capacity of 14,000 Btu/hr, whereas
the condensing unit selected has a capacity of
only 12,630 Btu/hr at the design suction
temperature of 20° F. Consequently, at the
design conditions, evaporator capacity will be
greater than condensing unit capacity, that is,
vapor will be produced in the evaporator at a
greater rate than it is removed from the evapo-
rator and condensed by the condensing unit.
Therefore, the system will not be in equilibrium
at these conditions. Rather, the excess vapor
will accumulate in the evaporator and cause an
increase in the evaporator temperature and
pressure. Since raising the evaporator tempera-
ture increases the suction temperature and, at
the same time, reduces the evaporator TD, the
condensing unit capacity will increase and the
evaporator capacity will decrease. System
equilibrium will be established when the
evaporator temperature rises to some point
where the suction temperature and evaporator
TD are such that the condensing unit capacity
and the evaporator are equal. In this instance,
system equilibrium, as determined graphically
(point A in Fig. 13-2), is established at a
suction temperature of approximately 23° F.
The evaporator TD is approximately 9° F,
which is 3° F less than the design evaporator
TD of 12° F and which will result in a space
humidity somewhat higher than the design
condition. The total system capacity is approxi-
mately 13,500 Btu/hr, which is about 23%
greater than the calculated hourly load of
11,000 Btu/hr. This means that the system
running time will be considerably shorter than
originally calculated. For instance, if the
original load calculation is based on a 16-hr
running time, the system will operate only
about 13 hr out of each 24.

The question immediately arises as to whether
or not this system will perform satisfactorily.
Although this would depend somewhat on the
particular application, the answer is that it
probably would not in the majority of cases.
There are several reasons for this. First, the
evaporator TD of 9° F is considerably less than
the design TD of 12° F and would probably

228 PRINCIPLES OF REFRIGERATION

| 22

§20

| 18

8 16

I"l4

&12

<"'

^\B

1 1 \

—

™

—

I —

1 \

0* 5* 10* 15* 20* 25'
Suction temperature

1 1 1 1 1 1

32* 27*

22* 17* 12*
Evaporator TD

Fig. 13-Z

result in a space humidity too high for the appli-
cation. Ordinarily, the humidity in the refrig-
erated space must be maintained within certain
fixed limits. Assuming that the design TD is
selected to produce the median condition within
these limits, a one degree deviation from the
design TD in either direction is usually the
maximum which can be allowed if the space
humidity is to be maintained within the limits
specified for the application.

Another consideration is the fact that the
system capacity is some 23% greater than the
calculated load so that the system operating
time will be relatively short. Although the sys-
tem capacity exceeds the calculated load by a
larger margin than good practice prescribes, this
in itself would not ordinarily cause any serious
problem in a majority of applications. How-
ever, since the shorter running time will also
tend to aggravate the already existing problem
of high humidity, especially in the wintertime,
when the two conditions are taken together, it
seems unlikely that the system would produce
satisfactory results in any application where the
space humidity is an important factor.

In the event that the equipment in question
represents the best available selection, the ques-
tion arises as to what can be done to bring the
system into balance at conditions more in
keeping with the design conditions. In this
instance, since the problem is one of excessive
evaporator capacity with relation to the con-
densing unit capacity at the design conditions,

logical corrective measures prescribe either an
increase in the condensing unit capacity or a
reduction in the evaporator capacity in order to
re-establish the point of system equilibrium at
conditions nearer to the design conditions.

Which of these two measures will produce the
most satisfactory results depends upon the rela-
tionship between the over-all system capacity
and the calculated load. Whereas increasing
either the condensing unit capacity or the evapo-
rator capacity will always bring about an in-
crease in the over-all system capacity, reducing
either the condensing unit capacity or the
evaporator capacity will always bring about a
reduction in the over-all system capacity.
Referring to Fig. 13-2 for the system under con-
sideration, if the condensing unit capacity is
increased to the evaporator capacity at the
design conditions, system equilibrium will shift
from point A to point B. On the other hand, if
the evaporator capacity is reduced to the con-
densing unit capacity at the design conditions,
the point of system equilibrium will shift from
A to C. Notice that, although the system is
balanced at the design conditions at either
points B or C, the over-all system capacity at
point B is considerably above the calculated
load, whereas at point C the over-all system
capacity very nearly approaches the calculated
load. Hence, in this instance, it is evident that
increasing the condensing unit capacity as a
means of bringing the system into balance at the
design conditions cannot be recommended,
since it would also increase the over-all system
capacity and therefore tend to aggravate the
already existing problem of excessive system
capacity with relation to the calculated load.
On the other hand, in addition to bringing the
system into balance at the design conditions,
reducing the evaporator capacity will also have
the beneficial effect of reducing the over-all
system capacity and thereby bringing it more
into line with the calculated load.
13-3. Decreasing or Increasing Evaporator
Capacity. Reducing the evaporator capacity
can be accomplished in several ways. One is to
"starve" the evaporator, that is, to reduce the
amount of liquid refrigerant in the evaporator
by adjusting the refrigerant flow control so that
the evaporator is only partially flooded with
liquid. This effectively reduces the size of the
evaporator, since that part of the evaporator

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 229

which is not filled with liquid becomes, in effect,
a part of the suction line.

Another method of reducing the capacity of
the evaporator is to reduce the air velocity over
the evaporator by slowing the evaporator fan or
blower. However, this method has its limita-
tions in that the air velocity must be maintained
at a level sufficient to assure adequate air
circulation in the refrigerated space. Too, re-
ducing the air quantity causes a change in the
sensible heat ratio of the evaporator. Depending
upon the particular application, this may or may
not be desirable.

As a general rule, there is little, if anything,
that can be done to increase the capacity of an
undersized evaporator. Occasionally, the evapo-
rator capacity can be increased by increasing
the air quantity. However, since increasing the
air quantity also increases air velocity and fan
horsepower requirements, this method has its
limitations for reasons already discussed in
Section 11-22.

In some cases, the evaporator surface area
can be increased somewhat by using a length of
eitherjjare tubing or finned tubing as a "drier
loop" or as additional evaporator surface. How-
ever, this too has its limitations because of the
pressure drop accruing in the tubing.
13-4. Decreasing or Increasing Condensing
Unit Capacity. Decreasing the condensing
unit capacity can be accomplished in several
ways, all of which involve decreasing the com-
pressor displacement. Probably the simplest
and most common method of reducing the con-
densing unit capacity is to reduce the speed of
the compressor by reducing the size of the pulley
on the compressor driver. The speed reduction
required is approximately proportional to the
desired capacity reduction.

The relationship between the speed of the
compressor and the speed of the compressor
driver is expressed in the following equation:

Rpm 1 x D x = Rpm t x D 2 (13-1)

where Rpm x = the speed of the compressor
(rpm)
D 1 = the diameter of the compressor
flywheel (inches)
Rpm t = the speed of the compressor
driver (rpm)
Z) 2 = the diameter of the driver pulley
(inches)

Note. Where the compressor driver is a
four-pole, alternating-current motor operating
on 60 cycle power, the approximate driver speed
is 1750 rpm. For a two-pole, alternating-current
motor, the approximate speed is 3500 rpm.

Example 13-2. A refrigeration compressor
having a 10 in. flywheel is driven by a four-pole,
alternating-current motor. If the diameter of
the motor pulley is 4 in., determine the speed
of the compressor.

Solution. Rearranging and
applying Equation 13-1, R x D

Rp mi = ^D,

_ 1750 x 4

10
= 700

Example 13-3. Determine the diameter of
the motor pulley required to reduce the speed
of the compressor in Example 13-2 from 700 to
600 rpm.

Solution. Rearranging and
applying Equation 13-1, D s

_ Rpm J x D t
Rpm^
600 x 10
1750
= 3.5 in.

Another method of reducing condensing unit
capacity is to reduce the volumetric efficiency of
the compressor by increasing the clearance
volume. This increase is accomplished by in-
stalling a thicker gasket between the cylinder
housing and the valve-plate.

In some cases, small increases in condensing
unit capacity can be obtained by merely in-
creasing the speed of the compressor. However,
when the capacity increase needed is substan-
tial, it is usually more practical and more
economical to use a larger size condensing unit
and reduce the capacity as necessary. The
reasons for this are several.

First, since the increase in compressor capa-
city will be accompanied by an increase in the
horsepower requirements, any substantial in-
crease in the compressor capacity will tend to
overload the compressor driver and necessitate
the use of a larger size. Too, some thought must
be given to the condenser capacity. Here again,
any increase in compressor capacity will tend to
place a heavier load on the condenser. If the
size of the condenser is not increased in propor-
tion to the increase in the condenser load,

230 PRINCIPLES OF REFRIGERATION

Fig. 13-3

excessive compressor discharge temperature and
pressure will result. Not only will this materially
reduce the life of the equipment and increase
maintenance and operating costs, but it will also
tend to nullify to some extent the gain in capa-
city originally accruing from the increase in
compressor speed.

It is apparent from the foregoing that, in most
cases, increasing the capacity of either the eva-
porator or the condensing unit is something
which is not easily accomplished. Therefore, it
is usually more practical and more economical
to select oversized equipment rather than under-
sized equipment. When the evaporator or the
condensing unit is oversized and capacity reduc-
tion is required to bring the system components
into balance at the desired conditions, the capa-
city reduction can readily be made with little, if
any, loss in system efficiency.
13-5. System Capacity vs. Calculated Load.
The relationship between system capacity and
system load is one which warrants careful con-
sideration and which can be best explained by
comparing the refrigerating system to a water
pumping system. For example, assume that it
is desired to maintain a constant water level in
the tank shown in Fig. 13-3. If the water flows
into the tank at a fixed and constant rate which
is readily computable, the water in the tank can
be maintained at a fixed level simply by installing
a pumping system which has a capacity exactly
equal to the flow rate of the water into the tank.
Since the flow rate of the water entering the tank
is constant and since the pumping rate is equal
to the water flow rate, the pump will operate
continuously and no other water level control of
any kind will be needed.

On the other hand, if the flow rate of the water
entering the tank varies from time to time, it is
evident that if the level of the water is to be
maintained within fixed limits, the pumping
system must be selected to have a capacity equal
to or somewhat in excess of the highest sustained
flow rate of the water entering the tank. It is
evident also that some means of cycling the
pump "off" and "on" must be provided. Other-
wise, during periods when the flow rate of the
water entering the tank is less than maximum,
the pumping rate will be excessive and the level
of the water in the tank will be reduced below
the desired levei. One convenient and practical
means of cycling the pump is to install a float
control in the tank (Fig. 1 3-4). The float control
is arranged to close the electrical contacts and
start the pump when the water in the tank rises
to a predetermined maximum level. When the
water level in the tank falls to a predetermined
lower limit, the float control acts to close the
electrical contacts and stop the pump. In this
way, intermittent operation of the pump will
maintain a relatively constant water level in
the tank.

The latter principle is readily applied to the
refrigerating system. Since the cooling load on a
refrigerating system varies from time to time,
the system is usually designed to have a capacity
equal to or somewhat in excess of the average
maximum cooling load. This is done so that the
temperature of the space or product can be
maintained at the desired low level even under
peak load conditions. As in the case of the water

To pump motor

Fig. 13-4

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 231

pumping system, since the system refrigerating
capacity will always exceed the actual cooling
load, some means of cycling the system "off"
and "on" is needed in order to maintain the
temperature of the space or product at a con-
stant level within reasonable limits and to pre-
vent the temperature of the space or product
from being reduced below the desirable
minimum.

For any refrigerating system, the relative
length of the "off" and "on" cycles will vary
with the loading of the system. During periods
of peak loading, the "running" or "on" cycles
will be long and the "off" cycles will be short,
whereas during periods of minimum loading the
"on" cycles will be short and the "off" cycles
will be long.*

13-6. Cycling Controls. The controls used to
cycle a refrigerating system "on" and "off" are
of two principal types: (1) temperature actuated
(thermostatic) and (2) pressure actuated. Each
of these types is discussed in the following
sections.

13-7. Temperature Actuated Controls.
Temperature actuated controls are called ther-

Bellows

<a)

Diaphragm ■~ >k .

(b)

Fig. 13-5. Bulb-type temperature sensing element.

* Unlike the water pumping system, refrigerating
systems are designed to have sufficient capacity to
permit "off" cycles even during periods of peak
loading. This is necessary in order to allow time for
defrosting of the evaporator. However, allowances
are made for defrosting time in the load calculations
(the 24-hr load, is divided by the desired running
time to obtain the average hourly load) and need
not be further considered when selecting the equip-
ment.

Stationary
contact

Bellows

Fig. 13-4. Schematic diagram of simplified pressure
control.

mostats. Whereas float controls are sensitive to
and are actuated by changes in liquid level,
thermostats are sensitive to and are actuated by
changes in temperature. Thermostats are used
to control the temperature level of a refrigerated
space or product by cycling the compressor
(starting and stopping the compressor driving
motor) in the same way that float controls are
used to control liquid level by cycling the pump
(starting and stopping the pump motor).
13-8. Temperature Sensing Elements. Two
types of elements are commonly used in thermo-
stats to sense and relay temperature changes to
the electrical contacts or other actuating mecha-
nisms. One is a fluid-filled tube or bulb which is
connected to a bellows or diaphragm and filled
with a gas, a liquid, or a saturated mixture of
the two (Figs. 13-5aand 13-56).t Increasing the
temperature of the bulb or tube increases the
pressure of the confined fluid which acts through
the bellows or diaphragm and a system of levers
to close electrical contacts or to actuate other
compensating mechanisms (Fig. 13-6). Decreas-
ing the temperature of the tube or bulb will
have the opposite effect.

f The thermostat described here is called a
remote-bulb thermostat. Although there are a
number of different types of thermostats, this is the
type most frequently used in commercial refriger-
ation applications. Thermostats are used for many
purposes other than controlling a compressor
driving motor, as, for example, opening and closing
valves, starting and stopping damper motors, etc.

232 PRINCIPLES OF REFRIGERATION

Dissimilar
metals

Invar-

___ Brass
"or steel

_Bimetal
"element

Normal
M

Fig. 13-7. Bimetal-t/pe tem-
perature sensing element.

Another and entirely different temperature
sensing element is the compound bar, commonly
called a bimetal element. The compound bar is
made up of two dissimilar metals (usually Invar
and brass or Invar and steel) bonded into a flat
strip (Fig. 13-7a). Invar is an alloy which has a
very low coefficient of expansion, whereas brass
and steel have relatively high coefficients of
expansions. Since the change in the length of
the Invar per degree of temperature change will
always be less than that of the brass or steel,
increasing the temperature of the bimetal ele-
ment causes the bimetal to warp in the direction
of the Invar (the inactive metal) as shown in
Fig. 13-76, whereas decreasing the temperature
of the bimetal element causes the bimetal to
warp in the direction of the brass or steel (the
active metal) as shown in Fig. 13-7c. This
change in the configuration of the bimetal ele-
ment with changes in temperature can be utilized
directly or indirectly to open and close electrical
contacts or to actuate other compensating
mechanisms.

13-9. Differential Adjustment. Like float
controls, thermostats have definite "cut-in" and
"cut-out" points. That is, the thermostat is
adjusted to start the compressor when the tem-
perature of the space or product rises to some
predetermined maximum (the cut-in tempera-
ture) and to stop the compressor when the
temperature of the space or product is reduced
to some predetermined minimum (the cut-out
temperature).

The difference between the cut-in and cut-
out temperatures is called the differential. In
general, the size of the differential depends upon
the particular application and upon the location

of the temperature sensing element. Where the
temperature sensing element of the thermostat is
located in or on the product and controls the
product temperature directly, the differential is
usually small (2° F or 3° F). On the other hand,
where the sensing element is located in the space
and controls the space temperature, the differen-
tial is ordinarily about 6° F or 7° F. In many
instances, the sensing element of the thermostat
is clamped to the evaporator so that the space
or product temperature is controlled indirectly
by controlling the evaporator temperature, in
which case the differential used must be larger
(15° F to 20° F or more) in order to avoid short-
cycling of the equipment.

When the thermostat controls the space or
product temperature directly, the average space
or product temperature is approximately the
median of the cut-in and cut-out temperatures.
Therefore, to maintain an average space tem-
perature of 35° F, the thermostat can be adjusted
for a cut-in temperature of approximately 38° F
and a cut-out temperature of approximately
32° F.

On the other hand, when the space tempera-
ture is controlled indirectly by controlling the
evaporator temperature, an allowance must be
made in the cut-out setting to compensate for
the evaporator TD. For example, for an
average space temperature of 35° F and assum-
ing an evaporator TD of 12° F, to compensate
for the evaporator TD, the cut-out temperature
would be set at 20° F (32° F - 12°) rather than
at 32° F. Notice that the cut-in temperature is
set at 38° F in either case. This is because the
space temperature and the evaporator tempera-
ture are the same at the time that the system

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 233

cycles on. After the compressor cycles off the
evaporator continues to absorb heat from the
space and warms up to the space temperature
during the ofT cycle (Fig. 13-8). Therefore, when
the space temperature rises to the cut-in tem-
perature of 38° F, the evaporator will also be at
the cut-in temperature of 38° F. As soon as the
compressor is started, the evaporator tempera-
ture is quickly reduced below the space tempera-
ture by an amount approximately equal to the
design evaporator TD. Therefore, in this in-
stance, when the space temperature is reduced
to 32° F (the desired minimum), the evaporator
temperature (which the thermostat is controlling)
will be approximately 20° F (12° F less than the
space temperature).

Regardless of whether the thermostat controls
the space temperature directly or indirectly,
proper adjustment of the cut-in and cut-out
temperatures is essential to good operation. If
the cut-in and cut-out temperatures are set too
close together (differential too small) the system
will have a tendency to short-cycle (start and
stop too frequently). This will materially reduce
the life of the equipment and may result in other
unsatisfactory conditions. On the other hand,
if the cut-in and cut-out temperatures are set too
far apart (differential too large), the on and off
cycles will be too long and unnecessarily large
fluctuations in the average space temperature
will result. Naturally, this too is undesirable.

Although approximate cut-in and cut-out
temperature settings for various types of appli-
cations have been determined by field experience,

in many cases it is necessary to use trial-and-
error methods to determine the optimum settings
for a specific installation.
13-10. Range Adjustment. In addition to the
differential, cycling controls have another adjust-
ment, called the "range," which is also asso-
ciated with the cut-in and cut-out temperatures.
Although, like the differential, the range can be
defined as the difference between the cut-in and
cut-out temperatures, the two are not the same.
For example, assume that a thermostat is
adjusted for a cut-in temperature of 30° F and a
cut-out temperature of 20° F. Whereas the
differential is said to be 10° F (30° - 20°), the
range is said to be between 30° F and 20° F.

Although it is possible to change the range
without changing the differential, it is not
possible to change the differential without
changing the range. For instance, suppose that
the thermostat previously mentioned is re-
adjusted so that the cut-in temperature is raised
to 35° F and the cut-out temperature is raised to
25° F. Although the differential is still 10° F
(35° — 25°), the operating range of the control
is 5° F higher than it was originally, that is, the
operating range is now between 35° F and 25° F,
whereas previously it was between 30° F and
20° F. In this instance, the range of the control
is changed, but the differential remains the same.
Under the new control setting the average space
temperature will be maintained approximately
5° F higher than under the old setting.

Suppose now that the differential is increased
5° F by raising the cut-in temperature from 30° F

-Average evaporator temperature

-Minimum evaporator temperature (cut-out temperature)

Fig. 13-8. Notice that when the unit cycles "on," the evaporator temperature is the same as the space
temperature, whereas when the unit cycles "off," the evaporator temperature is lower than the space
temperature by an amount equal to the design evaporator TD.

234 PRINCIPLES OF REFRIGERATION

, Differential adjustment

Fig. 13-9. Schematic diagram
of thermostatic motor control
illustrating range and differen-
tial adjustments.

Range
adjustment

to 35° F while the cut-out temperature is left at
the original setting of 20° F. Notice that both
the differential and the range are changed. The
differential, originally 10° F, is now 15° F and
the range, originally between 30° F and 20° F, is
now between 35° F and 20° F. With this control
setting, the running cycle will be somewhat
longer because the differential is larger. Too,
the average space temperature will be 2 or 3
degrees higher because the cut-in temperature is
higher. If the differential had been increased by
lowering the cut-out temperature 5° F rather
than by raising the cut-in temperature 5° F, the
operating range of the control would have
shifted to the opposite direction and the average
space temperature would have been 2 or 3
degrees lower than the original space tempera-
ture.

Typical range and differential adjustments
are shown in Fig. 13-9. Turning the range-
adjusting screw clockwise increases the spring
tension which the bellows pressure must over-
come in order to close the contacts and,
therefore, raises both the cut-in and cut-out
temperatures. Turning the range-adjusting
screw counterclockwise decreases the spring
tension and lowers both the cut-in and cut-
out temperatures.

Turning the differential-adjusting screw clock-
wise causes the limit bar A to move toward the
screw head, thereby increasing the travel of the
pin B in the slot. This has the effect of in-
creasing the differential by lowering the cut-out
temperature. Turning the differential adjusting
screw counterclockwise raises the cut-out tem-
perature and reduces the differential. By manipu-
lating both range and differential adjustments,
the thermostat can be adjusted for any desired
cut-in and cut-out temperatures.

The arrangement shown in Fig. 13-9 repre-
sents only one of a number of methods which
can be employed to adjust the cut-in and cut-out
temperatures. The particular method used in
any one control depends on the type of control
and on the manufacturer. For example, for the
control shown in Fig. 13-9, changing the range
adjustment changes both the cut-in and cut-out
temperatures simultaneously, whereas for an-
other type of control changing the range adjust-
ment changes only the cut-in temperature. For
still another type of control, changing the range
adjustment changes only the cut-out tempera-
ture. However, whatever the method of adjust-
ment, the principles involved are similar and the
exact method of adjustment is readily deter-
mined by examining the control . In many cases,

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 235

instructions for adjusting the control are given
on the control itself.

If electrical contacts are permitted to open or
close slowly, arcing will occur between the con-
tacts, and burning or .welding together of the
contacts will result. Therefore, cycling controls
which employ electrical contacts must all be
equipped with some means of causing the con-
tacts to open and close rapidly in order to avoid
arcing. In Fig. 13-9, the armature and per-
manent magnet serve this purpose. As the
pressure in the bellows increases and the
movable contact moves toward the stationary
contact, the strength of the magnetic field
between the armature and the horseshoe magnet
increases rapidly. When the movable contact
approaches to within a certain, predetermined,
minimum distance of the stationary contact, the
strength of the magnetic field becomes great
enough to overcome the opposing spring tension
so that the armature is pulled into the magnet
and the contacts are closed rapidly with a snap
action.

As the pressure in the bellows decreases,
spring tension acts to open the contacts. How-
ever, since the force of the spring is opposed
somewhat by the force of magnetic attraction,
the contacts will not separate until a consider-
able force is developed in the spring. This
causes the contacts to snap open quickly so that
arcing is again avoided.

Toggle mechanisms are also frequently used
as a means of causing the contacts to open and
close with a snap action. Too, some controls
employ a mercury switch as a means of over-
coming the arcing problem. A typical mercury
switch is illustrated in Fig. 13-10. As the glass
tube is tilted to the right, the pool of mercury
enclosed in the tube make contact between the
two electrodes. As the bulb is tilted back to the
left, contact is broken. The surface tension of
the mercury provides the snap action necessary
to prevent arcing.

13-11. Space Control vs. Evaporator Con-
trol. When the sensing element of the thermo-
stat is located in the space or in the product, the
thermostat controls the space temperature or
product temperature directly. Likewise, when
the sensing element is clamped to the evaporator,
the thermostat controls the evaporator tempera-
ture directly. In such cases, control of the space
or product temperature is accomplished in-

directly through evaporator temperature con-
trol. Which of these two methods of control is
the most suitable for any given application
depends upon the requirements of the applica-
tion itself.

For applications where close control of the
space or product temperature is desired, a
thermostat which controls the space or product
temperature directly will ordinarily give the best
results. On the other hand, for applications
where off-cycle defrosting is required and where
minor fluctuations in the space or product tem-
perature are not objectionable, indirect control
of the space temperature by evaporator tempera-
ture control is probably the better method.

In order to assure complete defrosting of the
evaporator, the evaporator must be allowed to
warm up to a temperature of approximately
37° or 38° F during each off cycle. When the
thermostat controls the evaporator temperature,
the cut-in temperature of the thermostat can be
set for 38° F. Since the evaporator must warm
up to this temperature before the compressor
can be cycled on, complete defrosting of the
evaporator during each off cycle is almost
certain. On the other hand, when the thermo-
stat controls the space temperature, there is no
assurance that the evaporator will always warm
up sufficiently during the off cycle to permit
adequate defrosting.

If we assume that the thermostat is properly
adjusted, if the load on the system is relatively
constant and the capacity of the system is
sufficient to handle the load, no defrosting prob-
lems are likely to arise with either method of
temperature control. However, if the system
is subject to considerable changes in load,
defrosting problems are sometimes experienced
in applications where the thermostat controls
the space temperature. When the system is

Glass tube

Pool of
mercury

Contacts
Fig. 13-10. Mercury contacts.

236 PRINCIPLES OF REFRIGERATION

operating tinder peak load conditions, the tem-
perature of the space tends to remain above the
cut-out temperature of the thermostat for
extended periods so that running cycles are long
and frost accumulation on the evaporator is
heavy. Too, under heavy load conditions, the
space temperature warms up to the cut-in
temperature of the thermostat very quickly
during the off cycle so that the off cycles are
usually short. Frequently, the off cycles are too
short to allow adequate defrosting of the eva-
porator. In such cases, when the compressor
cycles on again, the partially melted frost is
caught on the evaporator and frozen into ice.
Eventually the evaporator will be completely
frozen over with ice, air flow over the evaporator
will be severely restricted, and the system will
become inoperative.

13-12. Pressure Actuated Cycling Controls.
Pressure actuated cycling controls are of two
types: (1) low-pressure actuated and (2) high-
pressure actuated. Low-pressure controls are
connected to the low-pressure side of the system
(usually at the compressor suction) and are
actuated by the low-side pressure. High-
pressure controls, on the other hand, are con-
nected to the high-pressure side of the system
(usually at the compressor discharge) and are
actuated by the high-side pressure.

The design of both the low-pressure and the
high-pressure controls is similar to that of the
remote-bulb thermostat. The principal differ-
ence between remote-bulb thermostat and the
pressure controls is the source of the pressure
which actuates the bellows or diaphragm.
Whereas the pressure actuating the bellows of
the thermostat is the pressure of the fluid con-
fined in the bulb, the pressures actuating the
bellows of the low-and-high-pressure controls
are the suction and discharge pressures of the
compressor, respectively. Like the thermostat,
both controls have cut-in and cut-out points
which are usually adjustable in the field.
13-13. High-Pressure Controls. High-
pressure controls are used only as safety controls.
Connected to the discharge of the compressor,
the purpose of the high-pressure control is to
cycle the compressor off in the event that the
pressure on the high-pressure side of the system
becomes excessive. This is done in order to
prevent possible damaging of the equipment.
When the pressure on the high-pressure side of

the system rises above a certain, predetermined
pressure, the high-pressure control acts to break
the circuit and stop the compressor. When the
pressure on the high-pressure side of the system
returns to normal, the high-pressure control
acts to close the circuit and start the compressor.
However, some high-pressure controls are
equipped with "lock-out" devices which require
that the control be reset manually before the
compressor can be started again. Although
high-pressure controls are desirable on all
systems, because of the possibility of a water
supply failure, they are essential on systems
utilizing water-cooled condensers.

Since the condensing pressures of the various
refrigerants are different, the cut-out and cut-in
settings of the high-pressure control depend on
the refrigerant used.

13-14. Low-Pressure Controls. Low-pressure
controls are used both as safety controls and as
temperature controls. When used as a safety
control, the low-pressure control acts to break
the circuit and stop the compressor when the
low-side pressure becomes excessively low and
to close the circuit and start the compressor
when the low-side pressure returns to normal.
Like high-pressure controls, some low-pressure
controls are equipped with a lock-out device
which must be manually reset before the com-
pressor can be started.

Low-pressure controls are frequently used as
temperature controls in commercial refrigeration
applications. Since the pressure at the suction
inlet of the compressor is governed by the
saturation temperature of the refrigerant in the
evaporator, changes in evaporator temperature
are reflected by changes in the suction pressure.
Therefore, a cycling control actuated by changes
in the suction pressure can be utilized to control
space temperature indirectly by controlling the
evaporator temperature in the same way that
the remote-bulb thermostat is used for this
purpose. In such cases, the cut-in and cut-out
pressures of the low-pressure control are the
saturation pressures corresponding to the cut-in
and cut-out temperatures of a remote-bulb
thermostat employed in the same application.
For example, assume that for a certain appli-
cation the cut-in and cut-out temperature
settings for a remote-bulb thermostat are 38° F
and 20° F, respectively. If a low-pressure con-
trol is used in place of the thermostat, the cut-in

SYSTEM EQUILIBRIUM AND CYCLING CONTHOLS 237

pressure setting for the lew-pressure control wj|]
be SQpsia (the saturation pressure or R-12
corresponding to a temperature of 38 V} and
the cut-out pressure selling will be 36 psia (the
saturation pressure of R-12 corresponding to a
temperature or 20= F)-*

As thie evaporator warms up during the off
cycle, the pressure in the evaporator increases
accordingly. When the pressure in the evapora-
tor rises to the cut-in pressure setting of the low-
pressure control, the low-pressure control acts
to close the circuit and start the compressor.
Very soon after the compressor start*. the
temperature and pressure of ihe evaporator are
reduced to approximately the design evaporator
temperature and pressure and they remain nt
this condition throughout most of the running
cycle (see Fig. 13-8}, Near the end or the
running cycle the evaporator temperature and
pressure are gradually reduced below the design
conditions. When the evaporator pressure is
reduced to the cut-out pressure setting of the
low-pressure control, the control Ads to break
the circuit and stop the compressor.

Since the refrigerant vapor undergoes ■ drop
in pressure while flowing through the suction
line, the pressure of the vapor at the suction
inlet of the compressor is usually 2 or 3 lb less
than the evaporator pressure. This is particu-
larly true when the compressor is located
some distance from the evaporator. Since ihe
low-pressure control is actuated by the pressure
at the suction inlet of the compressor, the
pressure drop accruing in the suction line must
be taken into account when the pressure control
Bettings are made. To compensate for Ihe
pressure loss in the suction line, the cut-out
pressure setting is lowered by an amount equal
to the pressure loss in the lino. For example,
assuming a 3-1 b pressure loss in the suction line,
when the pressure in the evaporator is 36 psia,
the pressure at the suction inlet of the com-
pressor will be 33 psia. Hence, if it is desired to
cycle the compressor off when the pressure in
the evaporator is reduced to 36 psia, the cut-out
pressure of the low-pressure control is set for
33 psia. In this instance, failure to make an
nltowuiicc for the pressure loss in the Miction

* When refrigerants other than R-12 are raed ui
the system, the pressure setting! will be the m nar-
ration pr aam e s of those refrigerants corresponding
to the desired cut-in and cut-out temperature*

line would cause the control to cycle the com-
pressor when the pressure in the evaporator was
reduced to only 39 psia rather than the desired
36 psia. The system would have a tendency to
short cycle because the differential is too small
and unsatisfactory operation would result.

Pressure loss in the suction line in no way
ajfects the cut-in selling of the control. Since
pressure drop is a function of velocity or how,
there is no pressure drop in the suction line
when the system is idle. As soon as the com-
pressor cycles off, the pressure at the suction of
compressor rises to the evaporator pressure so
that at the time the compressor cycles on the
pressure at the compressor inlet is the same
as the evaporator pressure. Hence, the cut-in
pressure setting of the control is made without
regard for the pressure drop in the suction
line.

Since the low-pressure control controls eva-
porator temperature rather than space tempera-
ture, it is an ideal temperature control for
applications requiring off-cycle defrosting. This
is particularly true for '"remote" installations
where the compressor is located some distance
from the evaporator, tn such installations, low-
pressure temperature control has a distinct
advantage over thermostatic temperature con-
trol in that it ordinarily results in a considerable
saving in electrical wiring. Because of the
remote bulb, the thermostat must always be
located near the evaporator or space whose
temperature is being controlled. This requires
that a pair of electrical conductors be installed
between the fk)it are and the condensing unit. On
the other hand, the low-pressure control is
located at the compressor near the power source
so that the amount of control wiring needed is
much less.

13-15. Dual-Praiiur* Controls, A dual-pres-
sure control Is a combination or the low-and-
high- pressure controls in a single control.
Ordinarily, only one set of electrical contact
points are used in the control, although a
separate bellows assembly is employed for each
of the two pressures. A typical dual pressure
control is shown in Fig. 13-11. This type of
pressure control is frequently supplied as Stan*
dard equipment on condensing units.
13*16. The Pump-Down Cycle, A commonly
used method of cycling the condensing unit,
known as a "pump-down cycle," employs both a

33S PRINCIPLES OF REFRIGERATION

Rg. 11-11. Du*l pnmirt control, {Covrtmj 9< P«nn Contrail.

Inc.)

thermostat and a low-pressure control. In 1
pump-dawn cycle, the space or evaporator tccrh
perature is controlled directly by the thermostat.
However, instead of starting and stopping the
compressor driver, the thermostat nets 10 open
and clou a solenoid valve installed in the liquid
line, usually near the refrigerant flow control
(Fig. 13-12). As the space or evaporator tem-
perature is reduced to the cut-out temperature
of the thermostat, the thermostat breaks the
solenoid circuit, thereby de-energizing the
solenoid and interrupting the Gow of liquid
refrigerant to the evaporator- Continued
operation of the compressor causes evacuation
of the refrigerant from that portion of the
system beyond the point where the refrigerant
flow is interrupted by the solenoid. When the
pressure in the evacuated portion of the system
is reduced to [he cut-out pressure of the
low pressure control, the low-pressure control
breaks the compressor driver circuit and stops

the compressor. When the temperature of the
space or evaporator rises to tile cut-in tem-
perature of the thermostat, the thermostat
closes the solenoid circuit and energizes the
solenoid, thereby opening the liquid line and
permitting liquid refrigerant to enter the
evaporator. Since the c^ipumtar is warm,
the liquid entering the ■.-..ipuraujr vaporises
rapidly so that the evaporator pressure rises
immediately to the cut-in pressure of the
low-pressure control, whereupon the low-
pressure control closes the compressor driver
circuit and start* the compressor.

The advantages of the purip-down cycle are
many. One of the most important ones being
that the amount of refrigerant absorbed by the
oil in the cranfceasc of the compressor during the
olf cycle is substantially reduced. The problem
of crankcase oil dilution by refrigerant absorp-
tion during the off cycle is fully discussed in
Chapter IS.

STSTEM EQUILIBRIUM AND CYCLING CONTROLS 239

13-17. Variation i in System Capacity. Si is
worth while lo notice that balh the operating
conditions and the capacity ol" a refrigerating
system change as the load on I lie system changes.
When the load on the system is heavy and I he
space temperature is high, I he evaporator TD
will be somewhat larger than the design evapo-
rator TD and the capacity of the evaporator will
be greater than the design evaporator capacity.
Because of the higher evaporator capacity, the
suction temperature will also be higher so that
equilibrium is maintained between the vaporiz-
ing and condensing sections of the system.
Hence, under heavy load conditions, the system
operating conditions arc Somewhat higher than
the average design conditions and the system
capacity is somewhat greater than the average
design capacity. Obviously, the horsepower
requirements of the compressor are greatest
at this peak load condition and the compressor
driver must be selected lo have sufficient horse-
power lo meet these requirements.

Conversely, when the load on the system is
light, the space temperature wilt be lower (Kan
the average design space temperature, the
evaporator TD will be less than the design TD,
and the suction temperature will be lower than
the design suction temperature. Therefore, the
system operaiing conditions wilt be somewhat
tower than the average design operating con*
ditiofls and the system capacity will be some-
what less than the average design capacity

During each running cycle the system passes
through a complete series of operating condi-
tions and capacities, the operaiing conditions
and capacity being highest at the beginning of
the running cycle when the space temperature
is highest, and lowest at the end of the running
cycle when the space temperature Is lowest.
However, during most of the running cycle, a

well-designed system will operate very nearly
at the design conditions,

13-10, Capacity Control. The importance of
balancing the system capacity with the system
load cannot be overemphasized. Any time the
System capacity deviates considerably from the
system load, unsatisfactory operating conditions
will result. It has already been pointed out that
good practice requires that the system be
designed to have n capacity equal to or slightly
in excess of the average maximum sustained
load. This is done so that the system will have
sufficient capacity to maintain the temperature
and humidity at the desired level during
periods of peak loading. Obviously, as the
cooling load decreases, there is a lendency for
the system to become oversized in relation to
the load.

In applications where the changes in the
average system load are not great, capacity
control is adequately accomplished by cycling
the system on and oft as described in the
preceding sections. In such cases, assuming
that the cycling controls are property adjusted,
the relative length of the on and off cycles wilt
vary with the load on the system. During
periods when the load is heavy, on cycles will
be long and off cycles will be short, whereas
during periods when the load is light, on cycles
will be short and off cycles will be long. Natur-
ally, the degree of variation in the length of
the on and off cycles will depend on the degree
of load fluctuation. However, since the system
must always be designed to have sufficient
capacity to handle the maximum load, when
the changes in the system load are substantial,
it is evident that the system will be considerably
oversized when the load is at a minimum. A
system which is oversized for the load will
usually prove lo be as unsatisfactory as one that

!,„

flew antral

Flj. I S>-|1. Pump-down cycle.

I EvwofHw

hmmh i. :i i.

VPii

ftiwef JinT"

nwnmiii

140 PRINCIPLES OF MFMGERATtON

Fif, I J- 13. Evaporator iplit Into two lefmcnti lor
capacity control. Clining tht wl«noid rtlv» In (hi
llqwW tto* Of iff rtwfii A rtndtrt ihlt portion of
th* •viporvcor irwpemiva. Th* cipicn, reduction
obulnad U proportion*) to th* iiiriac* iru cydwl

is undersized for the load. When the system is
undersized for the load, the running: »"« will
be excessive, the space temperature will be high
Tor emended periods, and the off cycles will be
too short to permit adequate dcfrosiing of ihe
evaporator. On the other hand, where the
system is oversized for the load, the off cycles
will be loo long and the equipment running
time wtll he insufficient to remove the required
amount of moisture from the space. This will
result in unsatisfactory {higher than normal)
humidity conditions in the refrigerated space.

For this reason, when changes in the system
toad arc substantial, it is usually necessary to
provide some means of automatically (or
manually) varying the capacity of the system
other than by cycling Ihe system on and off.
This b true also of large installations when the
size of the equipment renders cycling the system
on and off impractical.

There are many methods of bringing the
refrigerating capacity Into balance with the
refrigerating load. Naturally, the most suitable
method in any one case will depend upon the
conditions and requirements of the installation
itself. Some installations require only one or
two step* of capacity control, whereas others
require a number of steps. Frequently, several
methods are employed simultaneously in order
to obtain the desired flexibility. Too, in some
cases, it is necessary to impose an artificial
load on the equipment to achieve the proper
balance between the sensible (temperature

reduction) and latent (nvuistun reduction)
loads.* In applications where the latent load
b too large a percentage of the total load,
satisfaction of the latent load will result in
ovcrcooling of the space utile** sensible heat is
artificially introduced into ihe space. In such
cases, the sensible heat is usually added to the
space in the form of relic, u The air is first
paased across a cooling coil and cooled to the
temperature necessary to reduce the moisture
content to the desired level ,md the air is then
reheated to ihe required dr\ hulb temperature.
The reheating is accomplished with steam or
hot water coils, with electric Mrip-heatcrs, or
with hot gas from the compressor discharge,

In some installations, iIk- refrigerating capa-
city of the system is adequately controlled by
controlling ihe capacity of the compressor only.
Since the flow rate of the refrigerant must be
the same in all component*, any change in the
capacity of any one component will automati-
cally result in a similar .iiljustment in the
capaci ly of al I the o ther com portents. Therefore,
increasing or decreasing the capacity of the
compressor will, in effect, increase or decrease
the capacity of the entire sisicm, However,
it is important to notice that with this method
of capacity control the operating conditions of
the system will change as the capacity of the
system changes.

Where it is desired to vnr\ ihe Capacity of the
System without allowing the operating condi-
tions of the system to change it is necessary lo

Fig- 11-14. Cell drtultad (or lac* control. (Court**?
Kifimrd Diviiion. Am»He»n Air Filter Company,
Inc.)

* This problem is usually rr,- - < < .it utc in the winter'
lime when the transmission i» .11 gAin) load is light.

SYSTEM EQUILIBRIUM AND CYCLING CONTROLS 341

control both the evaporator capacity and the
compressor capacity directly.*

Some of the more common methods of con-
trolling evaporator and compressor capacities
are considered in the following sections.
13-1?. Evaporator Capacity Control. Prob-
ably the most effective method of providing
evaporator capacity control it lo divide the
evaporator into several separate sections or
circuits which are individually controlled so that
one or more sections or circuit* an be cycled
out as the land decreases (Fig. 13-13}. Using
this method, any percentage of the evaporator
capacity con be cycled out in any desired
number of sieps. The number and size of
the individual evaporator sections depend on
the number of steps of capacity desired and the
percent change in capacity required per step,
respectively. The arrangement of the evaporator
sections or circuiting depends on the relation*
ship of the sensible load to ihe total load at 11k
various load conditions- Basically, two circuit
arrangements are possible- Evaporator circuit-
ing can be arranged lo provide cither "face"
control of "depth" control, or both (Fig. 13-14
and 11-15). When "face" control is used, the
"sensible heat ratio" of the evaporator is not

Hb> I J- IS, Coll circuit Ml for dapih control. (Cour*
teijr Ksrmind DMtl&n, American Air Filter Company,
Inc.).

* Any reduction in ivucjii loud unj/tit lyalcni
capacity will also have some elf«t on the capacity
of the condenser and on the sin of the refrigerant
lines. These topics lie d taunted in Chapter* J 4
and 17, respectively.

Fact damper

Fig. It 14, Evaporator equipped with * (ace damper
co vary the quantity otalrpasjin g ever chc evaporator.
Al the dam par movn toward the cloud paiition. the
rwiiunea aialrut which the Mower muat wsrk is
Increased so that the total quantity of air circulated

ofTected.t On the other hand, "depth" control
always changes the sensible heat ratio. As a
general rule, the more depth the evaporator has
the greater is its Jatcnt cooling (moisture
removal) capacity. Hence, as one or more rows
of the evaporator are cycled out, the sensible
heat ratio increases.

Another common method of varying the
evaporator capacity is to vary the amount of
air circulated over (he evaporator through the
use of "face" or "face-ond-by-pass" dampers.
(Fig*. 1 3- 16 and 13-17) Muflispeed blowers can
also be used for this purpose. Too, is some in-
siances, mul lis peed blowers and dampers are u%d
together if) order to provide the desired balance.

In nearly all cases, application of any of Ihe
foregoing methods of evaporator capacity
control will necessitate simultaneous control of
compressor capacity.

13-M, Compressor Capacity Control. There
in a number of different methods of controlling
the capacity of reciprocating compressors. One
method, already mentioned, is to vary the speed
of the compressor by varying the speed of the
compressor driver. When an engine or turbine
a employed to drive the compressor. Ihe
compressor capacity can he modulated over a
wide range by governor control of the com-
pressor driver.

When an electric motor drives the compressor.
Only two speeds are usually available SO that the
compressor operates either at full capacity or at
30% capacity. When more than two speeds ore
desired, it is necessary to use two separate
windings in the motor, in which case four speeds
will be possible.

; Ratio of the acntlbic coding capacity to the
total cooling capacity.

MI PWKCJPLES OF KEFRlGc RATION

By- pass damp**

! -

I -

Fk» dime**

Pit- fMT. Evaporator aqulpped with fit* and by-
pus dimptn lor dpicicy control. D*mp*r* if*
JrttereoAnectad w tint by-pui damper opum wider
a* ft« d»mp*r It tlowd eft. With thii iirinjerrmnt
tht quantity of air puling e*»r til* traporstor can
be rtfulated by allowing mart or leu »lr to by-pua
iht avipcrator . Hcwtver, ragirdilMJ of tba petition
of thi dim para, tht total quaintly of air circulated
remain* practically tht aim*.

Capacity control of multicylinder corn-
pretwrt it frequently obtained by "unloading"
one or more cylinders » that they become
ineffective. One method of accomplishing this
is to by-pas the discharge from one or more
cylinder] buck into the suction tine as shown
in Fig. 13-18. When the suction pressure drop*
to a certain predetermined value a solenoid
valve in the by-pass line, actuated by a pressure
control, opens and alky** ihe discharge from
one or more cylinders to flow through the
by-put tine back into the suction line where it
mixes with the incoming suction vapor. At
long as (he suction pressure remains below the
cut-in setting of the conirol. the discharge from
the unloaded cylinders continues to by-pus to
the suction line. When Ihe suction pressure
rises to the cut-out setting of the pressure-
con trot, the solenoid valve ,% dc-energixed and
the by-pass tine is closed so that the compressor
it returned to full capacity operation.

Another method of unloading compressor
cylinders is to depress the suction valves of the
cylinder or cylinders to be unloaded so that they
remain open during the compression (up)
stroke. With the auction v.thes held open, die
suction vapor drawn into the cylinder during
the suction stroke is returned to the suction line
during the compnuvDn stroke. A typical
unloadcr of this type it shown in Fig. 13-19,

The operation of the unloadcr mechanism is
as follows: when the suction pressure falls (o

the cut-in pressure of the prcuure control, the
control energizes the solenoid valve and admits
high-pressure gas from the condenser to the
unloadcr piston which act* to depress ihe
suction valves and bold them open. When
the suction pressure rises to the cut-out pressure
of the pressure control, the solenoid valve is de-
energized and the unloadcr piston is relumed to
the normal position.

In addition to providing; opacity control,
cylinder unloadcr* of all t;,pes art used to
unload the compressor cylinders during com-
pressor start-up so that the compressor starts
in an unloaded condition, thereby reducing (he
inrush current demand.

When any of the capacity control methods
described thus far are used. Ihe horsepower
requirements of the compressor decrease as the
capacity decreases, although not in the same
proportions.

Another method of controlling compressor
capacity is to throttle the cm if pressor suction.
However, since it reduces compressor capacity
uiihout reducing compressor horsepower, this
method it seldom used.

Still another meant of con trolling compressor
capacity which it employed with good results is
to operate two or more compressors in parallel
(Fig. 13-20). Individual low.. pressure controls
ore used to cycle the compressor*. The cut-in
and cut-out pressures of the mili vidua] controls
are so adjusted that the corn pressors cycle off
in sequence as the suction pressure decreases
and cycle on in sequence when the suction
pressure rises. Very often these compressors
are equipped wilh cylinder unio;iders to provide
additional steps of control Multiple com-
pressor systems are discussed in detail in
Chapter 20.

He. I ML Schauta** diagram of cyilAoV by-pan.

SYSTEM EQUILIBRIUM AND CYCUNG CONTROLS 243

Typical

solenoid valve

(shown energized)

Connect to
discharge side
of compressor

Fig. 13-19. Condenser pres-
sure actuated cylinder un-
loader mechanism. (Courtesy
Dunham-Bush, Inc.)

13-21. Multiple-System Capacity Control.

Another method of controlling capacity is to
employ two or more separate systems. The
evaporators for the separate systems may be in
the same housing and air stream or they may
be in separate housings and air streams. In
either case, separate compressors and condensers
are used, although in some instances the con-
densers may be in a common housing.

Fig. 13-20. Two compressors installed in parallel as
a means of controlling compressor capacity. As the
load diminishes, one compressor is cycled out to
reduce the compressor capacity. Often, one com-
pressor is equipped with cylinder unloaders to pro-
vide additional steps of control.

Valve plate
Cylinder

This method of capacity control is well
suited to installations where only two steps of
capacity control are required, as in chilling or
combination chilling and storage applications.
The use of two or more separate systems has the
added advantage of providing a certain amount
of insurance against losses accruing from
equipment failure. Should one system become
inoperative, the other can usually hold the load
until repairs can be made.

PROBLEMS

1. Assuming a 2° F loss in saturated suction
temperature due to refrigerant pressure drop
in the suction line.

(1) Select an air-cooled condensing unit to
operate in conjunction with the natural
convection evaporator in Problem 12-1.

(2) Plot the evaporator and condensing unit
capacities on a common graph and
determine:

(a) The saturated suction temperature at
the point of system balance

(b) The capacity of the system in Btu/hr at
the point of system balance.

Condensers and
Cooling Towers

14-1. Condensers. Like the evaporator, the
condenser is a heat transfer surface. Heat from
the hot refrigerant vapor passes through the
wails of the condenser to the condensing
medium. As the result of losing heat to the
condensing medium, the refrigerant vapor is
first cooled to saturation and then condensed
into the liquid state.

Although brine or direct expansion refrig-
erants are sometimes used as condensing
mediums in low temperature applications, in
the great majority of cases the condensing
medium employed is either air or water, or a
combination of both.

Condensers are of three general types:
(1) air-cooled, (2) water-cooled, and (3) evapo-
rative. Air-cooled condensers employ air as the
condensing medium, whereas water-cooled
condensers utilize water to condense the
refrigerant. In both the air-cooled and water-
cooled condensers, the heat given off by the
condensing refrigerant increases the temperature
of the air or water used as the condensing
medium.

Evaporative condensers employ both air and
water. Although there is some increase in
the temperature of the air passing through the
condenser, the cooling of the refrigerant in the
condenser results initially from the evaporation
of the water from the surface of the condenser.
The function of the air is to increase the rate of
evaporation by carrying away the water vapor
which results from the evaporating process.
14-2. The Condenser Load. Since the heat
given up by the refrigerant vapor to the con-

densing medium includes both the heat absorbed
in the evaporator and the heat of compres-
sion, the heat load on the condenser always
exceeds that on the evaporator by an amount
equal to the heat of compression. Since the
work (heat) of compression per unit of refrig-
erating capacity depends upon the compression
ratio, the heat load on the condenser per unit of
evaporator load varies with the operating
conditions of the system.

The quantity of heat liberated at the con-
denser per minute per ton of evaporator
capacity at various suction and condensing
temperatures can be estimated from Charts 14-1, .
14-2, and 14-3. Chart 14-1 applies to R-12
systems, whereas Charts 14-2 and 14-3 apply to
R-22 and R-717 (ammonia) systems, respec-
tively. The values given are based on a simple
saturated cycle.

Example 14-1. An R-12 system, operating
at a 15° F suction temperature, has a condens-
ing temperature of 100° F. Determine the load
on the condenser in Btu per minute per ton.

Solution. In Chart 14-1, locate the 15° F
suction temperature line at the base of the graph.
Follow the line until it intersects the 100° F
condensing temperature curve. The value on
the left-hand index corresponding to this point
is approximately 245 Btu per minute per ton.

It is evident that for any given set of operating
conditions there is a fixed relationship between
the condenser load and the evaporator load.
For instance, for the R-12 system described in
Example 14-1, the relationship between the
condenser load and the evaporator load is such
that 245 Btu are liberated at the condenser for
each 200 Btu taken in at the evaporator.

Once the relationship between the condenser
load and the evaporator load has been estab-
lished for any given set of operating conditions,
the total condenser load corresponding to any
given total evaporator load can be easily
computed. The following equation may be used :

ax*

(14-1)

where Q c = the condenser load in Btu/hr
Q e = the evaporator load in Btu/hr
q e = the condenser load in Btu/min/ton

(from Fig. 14-15)
q e = the evaporator load in Btu/min/ton

(always 200 Btu)

244

CONDENSERS AND COOLING TOWERS 245

Note: Q t may also be in Btu/min or in tons,
in which case Q e will be in Btu/min or in tons,
respectively.

Example 14-2. For the system described in
Example 14-1, determine the load on the con-
denser if the load on the evaporator is 35,000
Btu/hr.

Solution. Applying
Equation 14-1, the load = 35,000 x 245/200
on the condenser, Q c = 42,875 Btu/hr

It is important to notice that any increase or
decrease in the load on the evaporator (system)
will result in a proportional increase or decrease
in the load on the condenser.
14-3. Condenser Capacity. Since heat trans-
fer through the condenser walls is by conduction,
condenser capacity is a function of the funda-
mental heat transfer equation:

Q = A x U x D

(14-2)

where Q = the condenser capacity (Btu/hr)

A = the surface area of the condenser

(sqft)
U = the transfer coefficient of the con-
denser walls (Btu/hr/sq ft/° F)
D — the log mean temperature difference
between the condensing refrigerant
and the condensing medium
Examination of the factors in Equation 14-2
will show that for any fixed value of U
the capacity of the condenser depends on the
surface area of the condenser and on the
temperature difference between the condensing
refrigerant and the condensing medium. It is
evident also that for any one condenser of
specific size and design, wherein the surface
area and the U factor are both fixed at the time
of manufacture, the capacity of the condenser
depends only on the temperature differential
between the refrigerant and the condensing
medium. Therefore, for any one specific
condenser, the capacity of the condenser is
increased or decreased only by increasing or
decreasing the temperature differential. Further-
more, if it is assumed that the average tempera-
ture of the condensing medium is constant, it
follows that an increase or a decrease in the
capacity of the condenser is brought about only
by an increase or a decrease, respectively, in the
condensing temperature.

Since the condenser capacity must always be
equal to the condenser load, it is evident from
the foregoing that, for any given condensing
load, the larger the surface area of the condenser,
the smaller will be the required temperature
differential and the lower will be the condensing
temperature. Too, since the load on the
condenser is always proportional to the load on
the evaporator (system), any increase or decrease
in the load on the evaporator will be reflected by
an increase or a decrease, respectively, in the
condensing temperature.
14-4. Quantity and Temperature Rise of
Condensing Medium. In both the air-cooled
and water-cooled condensers, all the heat given
off by the condensing refrigerant increases the
temperature of the condensing medium. There-
fore, in accordance with Equation 2-8, the
temperature rise experienced by the condensing
medium in passing through the condenser is
directly proportional to the condenser load and
inversely proportional to the quantity and
specific heat of the condensing medium, viz:

(T a - T x ) =

M x C

(14-3)

where T x = the temperature of the air or water
entering the condenser (7",)
T 2 = the temperature of the air or water
leaving the condenser (T x )
(r 2 — Tj) = the temperature rise (AT) experi-
enced in the condenser
Q, = the load on the condenser in Btu

per hour
M = the weight of air or water circulated
through the condenser in pounds
per hour
C = the specific heat at constant
pressure of the air or water
Assuming that C has a constant value, for
any given condenser load (Q,), Equation 14-2
contains only two variables, M and AT, the
value of each being inversely proportional
to the value of the other, viz:

M =
AT =

C x AT
C xM

(14-4)
(14-5)

Therefore, for any given condenser load, if
the temperature rise of the condensing medium

246 PRINCIPLES OF REFRIGERATION

is known, the quantity of condensing medium
circulated through the condenser in pounds per
hour can be determined by applying Equation
14-4. Likewise, if the quantity circulated is
known, the temperature rise through the con-
denser can be computed by applying Equation
14-5.

Average specific heat values for air and water
are 0.24 Btu/lb and 1 Btu/lb, respectively. By
substituting the appropriate value for C,
Equations 14-4 and 14-5 can be written to apply
specifically to either water or air, viz:

(14-6)

(14-7)
(14-8)

(14-9)

Since general practice is to express air and
water quantities in cubic feet per minute (cfm)
and in gallons per minute (gpm), respectively,
it is usually desirable to compute condensing
medium quantities in these units rather than in
pounds per hour.

To convert pounds of water per hour into
gallons per minute, divide by 60 min to reduce
pounds per hour to pounds per minute, and
then divide by 8.33 lb per gallon to convert
pounds per minute to gallons per minute, viz:

for water

-£

»-%

and for air

M Q '

"' 0.24 x AT

IT Qs

*" 0.24 x M

gpm

M(lb/hr)

60 min x 8.33 lb/gal

If these conversion factors are incorporated
into Equation 14-6, the water quantity can be
computed directly in gpm. The following equa-
tion results:

Q.
8pm = 60 x 8.33 x Ar

or, combining constants (60 x 8.33

gpm

500 X AT

500),
(14-10)

convert from pounds per minute to cubic feet

per minute, viz:

M (lb/hr) x t>(cu ft/lb)

cfm = rz — :

60 min

Assuming the specific volume of the air to be
the specific volume of standard air (13.34 cu ft/
lb), incorporation of these conversion factors
into Equation 14-7 results in the following:
Q, x 13.34 cu ft/lb

To reduce pounds of air per hour to cubic
feet per minute, divide pounds per hour by 60
min to determine pounds per minute and then
multiply by the specific volume of the air to

cfm =

0.24 x 60 x AT

or, combining constants (13.34/0.24 x 60 =
1/1.08),

Qs (14-11)

cfm =

1.08.x AT

Example 14-3. If the load on a water-
cooled condenser is 150,000 Btu/hr and the
temperature rise of the water in the condenser
is 10° F. What is the quantity of water cir-
culated in gpm ?

Solution. Applying Equation
14-10, the water quantity in gpm

150,000

500 x 10

= 30 gpm

Example 14-4. TheJoad on a water-cooled
condenser is 90,000 Btu/hr. If the quantity of
water circulated through the condenser is 15
gallons per minute, determine the temperature
rise of the water in the condenser.

Solution. Rearranging and
applying Equation 14-10, AT

90,000
~ 500 x 15
= 12°F

Example 14-5. Thirty-six gallons of water
per minute are circulated through a water-
cooled condenser. If the temperature rise of the
water in the condenser is 12° F, compute the
load on the condenser in Btu/hr.

Solution. Rearrang-
ing and applying Equa- = 500 x Ar x gpm
tion 14-10, the load = 500 x 12 x 36
on Q, the condenser, = 216,000 Btu/hr

Example 14-6. The load on an air-cooled
condenser is 121,500 Btu/hr. If the desired
temperature rise of the air in the condenser is
25° F, determine the air quantity in cfm which
must be circulated over the condenser.

Solution. Applying Equa-
tion 14-11, the air quantity in
cfm

121,500
1.08 x 25
4500 cfm

CONDENSERS AND COOLING TOWERS 247

Fig. 14-1. Water temperature
rise through condenser.

92°-».Water out

Example 14-7. Three thousand cfm of air
are circulated over an air-cooled condenser. If
the load on the condenser is 64,800 Btu/hr,
compute the temperature rise of the air passing
over the condenser.

64,800

Solution. Rearranging and
applying Equation 14-1 1, AT = -^ -- ^

= 20°F

For any given condenser and condenser
loading, the condensing temperature of the
refrigerant in the condenser will depend only
upon the average temperature of the con-
densing medium flowing through the condenser.
The lower the average temperature of the con-
densing medium the lower is the condensing
temperature. For example, assume that the
size and loading of a condenser are such that
the required mean temperature differential
between the refrigerant and the condensing
medium is 15° F. If the average temperature of
the condensing medium is 90° F, the condensing
temperature will be 105° F (90 + 15), whereas if
the average temperature of the condensing
medium is 85° F, the condensing temperature
will be 100° F (85 + 15).

The average temperature of the condensing
medium flowing through the condenser depends
upon both the initial temperature of the
condensing medium entering the condenser and
the temperature rise experienced in the con-
denser. Since the temperature rise of the
condensing medium decreases as the flow rate
increases, the greater the quantity of condensing
medium circulated the lower is the average
temperature of the condensing medium. There-
fore, for any given condenser loading, the

greater the flow rate of the condensing medium
the lower will be the condensing temperature.
14-5. Condenser Application. As a general
rule, for any given condenser load, the size of
the condenser and the quantity of condensing
medium circulated will depend upon the
entering temperature of the condensing medium
and upon the desired condensing temperature.
A careful analysis of the data in Sections 14-3
and 14-4 will show that the condensing tempera-
ture of the refrigerant in the condenser is a
function of three variables: (1) the entering
temperature of the condensing medium, (2)
the temperature rise in the condenser, and (3) the
temperature difference between the refrigerant
and the condensing medium. This relationship
is illustrated in Fig. 14-1.

Recalling that the temperature rise in the
condenser varies inversely with the flow rate of
the condensing medium and that the tempera-
ture differential between the refrigerant and the
condensing medium varies inversely with the
size (surface area) of the condenser,* it is
evident that:

1. For any given condensing surface and
flow rate, die condensing temperature will
increase or decrease as the entering temperature
of the condensing medium increases or de-
creases.

2. For any given entering temperature, the
larger the condensing surface and the higher the
flow rate, the lower will be the condensing
temperature.

3. For any given entering temperature, the

* Assuming the transfer coefficient to be con-
stant.

248 PRINCIPLES OF REFRIGERATION

amount of condensing surface required for a
given condensing temperature decreases as the
flow rate of the condensing medium increases.

With regard to the latter statement, this
means in effect that the same condensing
temperature can be maintained with either a
small condensing surface and a high flow rate
or a large condensing surface and a low flow
rate. However, it should be recognized that the
flow rate of the condensing medium is fixed
within certain limits by the size and design of
the condenser. If the flow rate through the
condenser is too low, flow will be streamlined
rather than turbulent and a low transfer coeffi-
cient will result. On the other hand, if the flow
rate is too high, the pressure drop through the
condenser becomes excessive, with the result
that the power required to circulate the con-
densing medium also becomes excessive.

Since the design entering temperature of the
condensing medium is usually fixed by condi-
tions beyond the control of the system designer,
it follows that the size and design of the con-
denser and the flow rate of toe condensing
medium are determined almost entirely by the
design condensing temperature.

Although low condensing temperatures are
desirable in that they result in high compressor
efficiency and low horsepower requirements for
the compressor, this does not necessarily mean
that the use of a large condensing surface and a
high flow rate in order to provide a low con-
densing temperature will always result in the
most practical and economical installation.
Other factors which must be taken into account
and which tend to limit the size of the condenser
and/or the quantity of condensing medium
circulated are initial cost, available space, and
the power requirements of the fan, blower, or
pump circulating the condensing medium. Too,
where water is used as the condensing medium
and the water leaving the condenser is wasted
to the sewer (see Section 14-9), the availability
and cost of the water must also be considered.

The limitations imposed on condenser size by
the factors of initial cost and available space
are self-evident. As for the power requirements
of the fan, blower, or pump circulating the
condensing medium, it has already been stated
that the power required to circulate the con-
densing medium increases as the flow rate

increases. If the flow rate is increased beyond a
certain point, the increase in the power required
to circulate the condensing medium will more
than offset the reduction in the power require-
ments of the compressor accruing from the
increased flow rate. Therefore, the quantity of
condensing medium which can be economically
circulated is limited by the power requirements
of the fan, blower, or pump.

Obviously, the optimum flow rate for the
condensing medium is the one which will result
in the lowest over-all operating costs for the
system. This will vary somewhat with the
conditions of the individual installation, being
influenced by the type of application, the size
and type of condenser used, fouling rates, and
the design conditions for the region, along with
such practical considerations as the cost and
availability of water, utility costs, local codes
and restrictions, etc. For example, since good
system efficiency prescribes lower condensing
temperatures for low temperature applications
than for high temperature applications, it
follows that for the same condenser load the
optimum condensing medium flow rate will
usually be higher for a low temperature appli-
cation than for a high temperature application.
Too, where the entering temperature of the
condensing medium is relatively high, larger
condensing surfaces and higher flow rates
are required to provide reasonable condensing
temperatures than where the entering tempera-
ture of the condensing medium is lower.
14-6. Air-Cooled Condensers. The circula-
tion of air over an air-cooled condenser may be
either by natural convection or by action of a
fan or blower. Where air circulation is by
natural convection, the air quantity circulated
over the condenser is low and a relatively large
condensing surface is required. Because of
their limited capacity natural convection con-
densers are used only on small applications,
principally domestic refrigerators and freezers.

Natural convection condensers employed on
domestic refrigerators are usually either plate
surface or finned tubing. When finned tubing
is used, the fins are widely spaced so that little
or no resistance is offered to the free circulation
of air. Too, wide fin spacing reduces the
possibility of the condenser being fouled with
dirt and lint.

The plate-type condenser is mounted on the

CONDENSERS AND COOLING TOWERS 249

back of the refrigerator in such a way that an
air flue is formed to increase air circulation.
Finned tube condensers are mounted either on
the back of the refrigerator or at an angle
underneath the refrigerator. Regardless of
condenser type or location, it is essential that the
refrigerator be so located that air is permitted
to circulate freely through the condenser at all
times. Too, warm locations, such as one
adjacent to an oven, should be avoided when-
ever possible.

A number of domestic freezer manufacturers
utililize the outer shell of the freezer (outside
wall surface) as a condensing surface. This is
accomplished by bonding bare tubing to the
inside surface of the outer shell so that the
entire outer shell becomes a plate-type heat
transfer surface. The use of these "wrap-
around" condensers permits a considerable
reduction in the size of the freezer (6 to 8 in. on
both length and width), not only because it
eliminates the space ordinarily required for the
condenser but also because it allows the use of
3 to 4 in. of insulation in the walls where
normally 6 to 8 in. is required in order to prevent
moisture from condensing on the outside surface
of the freezer. The slightly higher operating
costs which accrue as a result of reducing the
amount of wall insulation is more than offset
by the savings in space that this practice makes
possible.

Air-cooled condensers employing fans or
blowers to provide "forced-air" circulation can
be divided into two groups according to the
location of the condenser: (1) chassis-mounted
and (2) remote.

A chassis-mounted air-cooled condenser is
one that is mounted on a common chassis with
the compressor and compressor driver so that
it is an integral part of the air-cooled "con-
densing unit" (Fig. 6-12). Although chassis-
mounting of the air-cooled condenser makes
possible a very compact, completely self-
contained condensing unit which is ideally
suited for use on small commercial fixtures,
this arrangement has certain inherent disadvan-
tages which make chassis-mounting impractical
in larger applications.

The principal disadvantage of chassis-
mounted air-cooled condensers is that the
physical size of the condenser is limited to
the dimensions of the chassis. Because of the

limitation in physical size, chassis-mounted
condensers of the type shown in Fig. 6-12 are
rarely found in capacities larger than 2 tons.*

Another disadvantage of the chassis-mounted
air-cooled condenser is their susceptibility to
fouling. Since most condensing units are
mounted on the floor, the condenser air tends
to sweep across the floor so that dirt, lint, and
other foreign materials are picked up from the
floor and carried to the surface of the condenser,
thereby "clogging" the condenser and restricting
the air flow.

Too, on "open-type" air-cooled condensing
units the condenser fan is usually mounted on
the shaft of the compressor driver (Fig. 6-12).
Naturally this limits both the size and the
location of the fan so that the quantity of air
circulated over the condenser is always less
than that which would produce maximum
efficiency at full load conditions. Notice also
that, because of the fan location, the distri-
bution of the air over the condenser surface is
very uneven, being much greater on the end of
the condenser directly in front of the fan.

Remote air-cooled condensers are used in
sizes from 1 ton up to 100 tons or more and may
be mounted either indoors or outdoors. When
located indoors, provisions must be made for an
adequate supply of outside air to the condenser
(Fig. 14-2). If the condenser is installed in a
warm location, such as in an attic or boiler
room, ducts should be used to carry the air into
the condenser and back to the outside. Because
of the large quantity of air required, only the
smaller sizes are mounted indoors.

When located outdoors, the air-cooled con-
denser may be mounted on the ground, on the
roof, or on the side of a wall, with roof locations
being the most popular. Typical wall and roof
installations are shown in Figures 14-3 and
14-4, respectively. In all cases, the condenser
should be so oriented that the prevailing winds
for the area in the summertime will aid rather
than retard the action of the fan. In the event
that such orientation is not possible, wind
deflectors should be installed on the discharge
side of the condenser (Fig. 14-5).

* This is the approximate condenser capacity
required on a 3 hp, commercial, air-cooled con-
densing. Approximately 25 % of the motor horse-
power is required to drive the fan. Naturally, this
reduces the horsepower available to the compressor.

250 PRINCIPLES OF REFRIGERATION

Ceiling

Purge valve

Fig. 10-2. Indoor installation of
air-cooled condenser. (Cour-
tesy Kramer Trenton Com-
pany.)

/ Compressor''

* Locate receiver below unicon outlet

One significant outgrowth of the remote air-
cooled condenser has been the development of
a new type of air-cooled condensing unit which
is designed specifically for remote installation.
The air-cooled condensing unit illustrated in
Fig. 14-6 is typical of these newer designs. This

type is rapidly gaining in popularity and is now
available in almost any desired capacity.
14-7. Air Quantity and Velocity. For an
air-cooled condenser there is a definite relation-
ship between the size (face area) of the con-
denser and the quantity of air circulated in that
the velocity of the air through the condenser is
critical within certain limits. Good design pre-
scribes the minimum air velocity that will pro-
duce turbulent flow and a high transfer coefficient.
Increasing the air velocity beyond this point
causes an excessive pressure drop through the
condenser and results in an unnecessary increase
in the power requirements of the fan or blower
circulating the air.

The velocity of the air passing through an air-
cooled condenser is a function of the free face
area of the condenser and the quantity of air
circulated. The relationship is given in the
following equation:

Air velocity (fpm) =

Air quantity (cfm)
Free face area (sq ft)

Fig. 14-3. Remote air-cooled condensers installed on
outside wall. (Courtesy Kramer Trenton Company.)

The free face area of the condenser is the area
of the free air spaces between the tubes and fins.
The actual free area per unit of face area varies
with the design of the condenser, being depend-
ent upon the size, number, and arrangement of
the tubes and fins.

CONDENSERS AND COOLING TOWERS 25 1

Fig. 14-4. Remote air-cooled condensers mounted on roof. (Courtesy Dunham-Bush, Inc.)

Normally, air velocities over air-cooled con-
densers are between 500 and 1000 fpm. How-
ever, because of the many variables involved,
the optimum air velocity for a given condenser
design is best determined by experiment. For
this reason, most air-cooled condensers come
from the factory already equipped with fans or
blowers so that the air quantity and velocity
over the condenser are fixed by the manu-
facturer. In all cases, to realize peak perform-
ance from an air-cooled condenser, the

manufacturer's recommendations as to the air
quantities should be carefully followed.
14-8. Rating and Selection of Air-Cooled
Condensers. Capacity ratings for air-cooled
condensers are usually given in Btu/hr for
various operating conditions. It has already
been shown that since the surface area and the
value of U are fixed at the time of manufacture,
the capacity of any one condenser depends only
on the mean temperature difference between
the air and the condensing refrigerant. Since

Fig. 14-5. Remote air-cooled
condensers equipped with
wind deflectors. (Courtesy
Kramer Trenton Company.)

252 PRINCIPLES OF REFRIGERATION

Fig. 14-6. Air-cooled condensing unit designed for remote installation. Notice generous size of condenser.
(Courtesy Kramer Trenton Company.)

most air-cooled condensers come equipped with
fans or blowers, the quantity of air circulated
over the condenser is also fixed so that the
average temperature of the air passing over the
condenser depends only on the dry bulb tem-
perature of the entering air and the load on the
condenser. Obviously, then, in such cases, the
capacity of the condenser is directly proportional
to the temperature difference between the dry
bulb temperature of the entering air and the
condensing temperature. This temperature
differential is often referred to as the "tempera-
ture split" in order to distinguish it from the
mean effective temperature differential.*

Table R-13 is a typical manufacturer's rating
table for air-cooled condensers. The basic
ratings given in Table R-13A are based on 90° F
ambient air temperature, 120° F condensing
temperature, and 40° F evaporating tempera-
ture. For other design conditions multiply the
basic rating from Table R-13 A by the correction
factors for variation in evaporating temperatures
(Table R-13B) and for variation in entering air
and condensing temperatures (Table R-13Q.

* The temperature split is always proportional to
the METD.

In order to select a condenser from the rating
tables, the following design data must be known :

(1) The design suction and condensing tem-
peratures

(2) The compressor capacity in Btu/hr

(3) The design outdoor dry bulb temperature
(use values in Table 10-6A. Round off to
next highest multiple of 5)

Example 14-8. From Table R-13, select
an air-cooled condenser for a compressor
having a capacity of 75,000 Btu/hr if the design
evaporating and condensing temperatures are
20° F and 110° F, respectively, and the outdoor
design dry bulb for the region is 90° F.

Solution. From Table
R-l 3B, the correction fac-
tor for 20° F suction tem-
perature = 0.95

From Table R-12C, the
correction factor for con-
densing temperature of
110° F and entering air
temperature of 90° F = 0.665

Required capacity of
condenser at basic rating 75 000

conditions - 0.95 x 0.665

= 11 8,700 Btu/hr

CONDENSERS AND COOLING TOWERS 253

From Table R-13A, select condenser Model
#BD1000 which has a capacity of 120,000
Btu/hr at the basic rating conditions.

Experience has shown that as a general rule
selecting an air-cooled condenser on the basis
of a condensing temperature of 1 10° F will
result in the most economical condenser size.
Hence, the actual size of the condenser selected
will depend upon the outdoor design dry bulb
temperature for the region in question. The
higher the dry bulb temperature, the larger the
condenser required. For example, for a con-
densing temperature of 1 10° F, if the dry bulb
temperature is 85° F, the condenser can be
selected for a 25° F temperature split, whereas
if the dry bulb temperature is 90° F, the con-
denser must be selected for a 20° F temperature
split, which will require a larger size.
14-9. Water-Cooled Condenser Systems.
Systems employing water-cooled condensers can
be divided into two general categories: (1)
waste-water systems and (2) recirculated water
systems. In waste-water systems the water
supply for the condenser is usually taken from
the city main and wasted to the sewer after
passing through the condenser (Fig. 14-7). In
recirculated water systems the water leaving the
condenser is piped to a water cooling tower
where its temperature is reduced to the entering
temperature, after which the water is recircu-
lated through the condenser (Fig. 14-8).

Naturally, where the condenser water is
wasted to the sewer, the availability and cost of
the water are important factors in determining
the quantity of water circulated per unit of con-
denser load. As a general rule, an economical
balance between water and power costs pre-
scribes a water flow rate of approximately 1.5
gal per minute per ton of capacity.

The high cost of water, along with limited
sewer facilities and recurring water shortages in
many regions, has tended to limit waste-water
systems to very small sizes. Too, many cities
have placed severe restrictions on waste-water
systems, particularly where the water supply is
taken from the city main and wasted to the
sewer.

When the condenser water is recirculated the
power required to circulate the water through
the water system must be taken into account in
determining the water flow rate. Experience has

shown that, in general, a water flow rate of
between 2.5 and 3 gal per minute per ton usually
provides the most economical balance between
compressor horsepower and pump horsepower.

In some instances, the water supply for a
waste-water system is taken from a well or from
some nearby body of water, such as a river,
lake, pond, etc., in which case both the cost of
the water and the pumping horsepower must be
considered in determining the optimum water
flow rate.

To a large extent, the quantity of water cir-
culated through the condenser determines the
design of the water circuit in the condenser.
Since heat transfer is a function of time, it
follows that where low water quantities necessi-
tate a high temperature rise in the condenser,
the water must remain in contact with the con-
densing refrigerant for a longer period than
when the water flow rate is high and the tem-
perature rise required is smaller. Hence, where
the water flow rate is low, the number of water
circuits through the condenser are few and the
circuits are long so that the water will remain in
the condenser for enough time to permit the
required amount of heat to be absorbed. On
the other hand, when the flow rate is high and
the temperature rise low, more circuits are used
and the circuits are shorter in order to reduce
the pressure drop to a minimum. This is illu-
strated in Figs. 14-9a and 14-96. In Fig. 14-9a,
the two water circuits through the condenser are
connected in series for a low flow rate and a high
temperature rise. The water enters through
opening A and leaves through opening C.
Opening B is capped. In Fig. 14-9ft, the two
water circuits are connected in parallel for a high
flow rate and a low temperature rise. The water
enters through opening B and leaves through
openings A and C.

In designing the condenser water circuit
particular attention must be given to the water

Suction

/Water regulating valve
Warm mater out _^.

Fig. 14-7. Waste water system.

254 PRINCIPLES OF REFRIGERATION

Hot gas in

Fig. 14-8. Recirculating water
system.

Pump-^

velocity and pressure drop through the con-
denser. In all cases the minimum permissible
velocity is that which will produce turbulent
flow and a high transfer coefficient. Since
pressure drop is a function of velocity, the
pressure drop through the condenser increases
as the water velocity increases. For this reason,
the maximum permissible velocity in any one
case is usually determined by the allowable
pressure drop. * For waste-water systems, where

to)

Water out

//t\rtsJDhJjt\

C Water in

| *■* >* ** '*'

~U A\ A\ A\ A\

\y \y \y_\y w

A Water in

<b)

Fig. 14-9. (o) Water circuit connected for series
flow, (b) Water circuit connected for parallel flow.

* Excessive velocity will usually cause erosion of
the water tubes, particularly at points where the
water changes direction. The maximum velocity
recommended by Air Conditioning and Refrigeration
Institute (ART) is 8 fps.

the water is forced through the condenser by
city main pressure, the pressure drop through
the condenser is not critical as long as it is
within the limits of turbulent flow and the avail-
able city main pressure. In such cases, high
velocities are recommended in order to take
advantage of the higher transfer coefficient. On
the other hand, when the water is circulated by
action of a pump, a high pressure drop through
the condenser will increase the pumping head
and the power required to circulate the water.
Therefore, for recirculating water systems, the
optimum water velocity is one which will pro-
vide the most economical balance between a
high transfer coefficient and a low pumping
head.

In Figs. 14-9a and 14-9A, it is of interest to
notice that for the same flow rate the velocity
and pressure drop through the circuit arrange-
ment in Fig. 14-9a are approximately four times
as great as that through the circuit arrangement
in Fig. 14-96. Too, because of the higher velo-
city, the transfer coefficient is somewhat higher
for the condensing surface in Fig. 14-9a and less
condensing surface is required for the same heat
transfer capacity.

14-10. Fouling Rates. Another factor which
must be considered in selecting a water-cooled
condenser is fouling of the tube surface on the
water side. The fouling is caused primarily by
mineral solids which precipitate out of the water
and adhere to the tube surface. The scale thus
formed on the tube not only reduces the water
side transfer coefficient, but it also tends to

CONDENSERS AND COOLING TOWERS 255

restrict the water tube and reduce the quantity
of water circulated, both of which will cause
serious increases in the condensing pressure.

In general, the rate of tube fouling is in-
fluenced by: (1) the quality of the water used
with regard to the amount of impurities con-
tained therein, (2) the condensing temperature,
and (3) the frequency of tube cleaning with
relation to the total operating time.

Most manufacturers of water-cooled con-
densers give condenser ratings for clean tubes
and for four stages of tube fouling in accordance
with the scale factors given in Table 14-1 for
various types of water. These scale factors are
an index of the reduction in the tube transfer
coefficient resulting from the scale deposit. In
selecting a water-cooled condenser, a minimum

Refrigerant
" vapor in

this arrangement, some air-cooling of the refrig-
erant is provided in addition to the water-
cooling. Counterflowing of the fluids in any
type of heat exchanger is always desirable since
it results in the greatest mean temperature
difference between the fluids and, therefore, the
highest rate of heat transfer.

Several types of double-tube condensers are
shown in Figs. 14-11 and 14-12. The type
shown in Fig. 14-11 can be cleaned mechan-
ically by removing the end-plates (inset). The
type shown in Fig. 14-12 is cleaned by circu-
lating approved chemicals through the water
tubes (see Section 14-23).

Equipped with water-regulating valves (Sec-
tion 14-20), double-tube condensers make ex-
cellent "booster" condensers for use with

Water out

Fig. 14-10. Double -tube
water-cooled condenser.

Water in^

scale factor of 0.0005 should always be used.
Under no circumstances should a condenser be
selected on the basis of clean tubes. However,
when the condensing temperature is low (leaving
water temperature less than 100° F) and the
condenser tubes are to be cleaned frequently,
the fouling factor from Table 14-1 may be
reduced to the next lowest value. The use of
scale factors will be illustrated later in the
chapter.

14-11. Water-Cooled Condensers. Water-
cooled condensers are of three basic types:
(1) double-tube, (2) shell-and-coil, and (3) shell-
and-tube.

As its name implies, the double-tube con-
denser consists of two tubes so arranged that
one is inside of the other (Fig. 14-10). Water is
piped through the inner tube while the refrig-
erant flows in the opposite direction in the
space between the inner and outer tubes. With

Condensed
refrigerant out

chassis-mounted air-cooled condensers during
periods of peak loading. Since the water valve
can be adjusted to open and allow water to flow
through the condenser only when the con-
densing pressure rises to some predetermined
level, the amount of water used is relatively
small in comparison to the savings in power
afforded by the increased compressor effi-
ciency.

The shell-and-coil condenser is made up of
one or more bare-tube or finned-tube coils
enclosed in a welded steel shell (Fig. 14-9). The
condensing water circulates through the coils
while the refrigerant is contained in the shell
surrounding the coils. Hot refrigerant vapor
enters at the top of the shell and condenses as it
comes in contact with the water coils. The con-
densed liquid drains off the coils into the bottom
of the shell, which often serves also as the
receiver tank. Care should be taken not to

256 PRINCIPLES OF REFRIGERATION

Fig. 14-11, Double-pipe condenser* wlih mechanically deanable tube*. {Courtwy Hilnead and Mitchell.)

overcharge the system with refrigerant since an
excessive accumulation of liquid in the ton-
denser will tend to cover too much of the con-
densing surface and cause an increase in Frit
discharge temperature and pressure.

Most sheli-and-coil condensers are equipped
with a split water circuit. The two parts of the
circuit are connected in series for waste-water
systems (Fig. J 4-96) and in parallel for recircu-
lating systems (Fig. i4-9a). As a general rule,
shcll-and-coil condensers arc used only for small
installations up to approximately 10 tons
capacity,

Shelt-and-coil condensers are cleaned by cir-
culating an approved chemical through the
water coils.

The sbcll-and-tube condenser consists of a
cylindrical steel shell in which a number of
straight tubes are arranged in parallel and held
in place at the ends by tube sheets. Construc-
tion is almost identical to that of the flooded-
type shell-and-tube liquid chiller. The con-
densing water is circulated through the tubes,
which may be either steel or copper, bare or
extended surface. The refrigerant is contained
in the steel shell between the tube sheets. Water
circulates in the annular spaces between the tube
sheets and the end-piates, the end-plates being
baffled to act as manifolds to guide the water
flow through the tubes. The arrangement of the
end-plate bathing determines the number of
passes the water makes through the condenser

from one end to the other before leaving the
condenser. The number of passes may be as few
as two or as many as twenty.

For any given total number of lubes, the
number of tubes per pass varies inversely with
the number of passes. For example, assuming
that a condenser has a total of forty tubes, if
there are two passes, the number of tubes per
pass is twenty, whereas if there are four passes,
the number of tubes per pass is ten.

It is important to notice that for the same
total number of tubes and the same water
quantity, the velocity of the water and the
pressure drop through the condenser will be four
times as great for a four-pass condenser as for a
two-pass condenser. Because of the higher velo-
city the transfer coefficient will be higher for the
four-pass condenser and a smaller condensing
surface will be required for a given heat transfer
capacity. However, on the other hand, because
of the high pressure drop, the power required to
circulate the water will be greater. Hence, for a
waste- water system, the- four-pass condenser is
probably the best selection, whereas for a
recirculating system, the two-pass condenser is
probably the better of the two.*

Shcil-and-fube condensers are available in
capacities ranging from 2 tons up to several

* This example is intended only In illustrate the
principles of design and should not be construed
to mean that four-pass condensers are undesirable
for recirculating systems.

CONDENSERS AND COOLING TOWERS 157

Hg. 14-12. Typical double-pipe condenser cortfigur«iofl*. (fl) Trombone configuration, (t) HeUlt configura-
tion. (Courtesy Edwardi Engineering Corporation.)

hundred tons or more. Shell diameters range
from approximately 4 in, up to 60 in,, whereas
tube length varies from approximately j ft to 20
ft. The number and the diameter or the tubes
depend on the diameter of the shell. Tube
diameters of g in. through 2 in. are common,
whereas the number of Lubes in the condenser
varies from as few as six or eight to as many as a
thousand or more. The end-plates of the con-
denser are removable to permit mechanical
cleaning of the water tubes.

Single-pass, vertical sheH-and-tube condensers
are sometimes employed on large ammonia
installations. The construction of the vertical
she! I -and- lube condenser is similar to that of the
vertical shcll-and-tubc chiller illustrated in Fig.
1 1 -44. The vertical condenser is equipped * i th a
water box at the top to distribute the water to
the tubes and a drain at the bottom to carry the
water away. Bach tube is equipped at the top
with a distributor fitting which imparts a rotating
motion to the water to assure adequate wetting
of the lube. The hot refrigerant vapor usually
enters at the side of the shell near the middle of
the condenser and the liquid leaves the con-
denser at the side of the shell near the bottom.
The height of vertical shell-and-tube condensers
ranges from 12 ft to IE ft. The tubes are
mechanically clcanabk.

14-12- Rating and Selection of Water'
Cooled Condensen.* The ratings shown in

* The maLcriiil in this section is reprinted directly
from the manufacturer's catalog, the only alter*
ation being the. table designations. Courtesy of
Acme Industries, Inc.

Table R-I4 are based on condensing tempera-
tures of 102° and 105^ F, 20" and 10" water rise
and 0.0005 scale factor which is the minimum
recommended in ARf standards.

Where other conditions exist, the following
procedure should be followed in selecting the
proper condenser.

Condensers must not be selected for less than
0.5 gpm per tube below which streamline instead
of turbulent water flow occurs. ART standards
indicate that the water velocity should not
exceed E fps which is 5.75 gpm per tube for
Acme STF and SRF condensers.

It is necessary to have the following informa-
tion to select a proper condenser :

1. Total tons (low side).

2. Evaporator temperature.

3. Condensing temperature.

4. Water temperature "in."

5. Water temperature "out," or gpm avail-
able.

6. Type of water or required scale factor.

Then proceed as follows :

1 . Determine the corrected tons to be used in
selecting the proper condenser by reference to
Fig. 2, Table R-I4. The factor obtained for the
desired evaporator temperature and condensing
temperature is multiplied by the actual tons to
obtain corrected tons.

2. Determine the water temperature rise and
gpm per ton. Knowing cither factor, the other
may be obtained by reference to Fig. 3, Table
R-14. Use corrected tons to determine the total
gpm required.

258 PRINCIPLES OF REFRIGERATION

3. Determine the temperature differences
between the condensing temperature and the
"water in" and "water out" temperatures and
find the METD by referring to Table 11-1.

4. Make preliminary selection of condenser
shell diameter by reference to Table R-14,
basing the selection on the corrected tons found
in step 1 . Find the number of tubes per pass and
then by referring to step 2, find the gpm per
tube.

5. Select the desired scale factor by reference
to Table 14-1 which suggests scale factors for
various types of water. The most commonly
used factor is 0.0005 and it should be borne in
mind when selecting a factor that a determina-
tion is being made of the frequency of cleaning
which will be required.

6. Referring to Fig. 1, Table R-14, determine
the rate of heat transfer "£/" for the gpm per
tube in step 4 and the scale factor in step 5.

7. Calculate the surface required by use of the
following formula.

Square feet of surface

Corrected tons x 14,400
= U x METD

8. Select a condenser having at least the
required surface from Table R-14. Be sure to
use the shell diameter determined in the pre-
liminary selection of step 4.

9. Make final checks on selection.

a. Using the gpm per tube from step 4 and
the nominal tube length shown in Table R-14
for the model selected in step 8, refer to Fig.
4 of Table R-14 to obtain water pressure drop
through condenser.

b. Obtain nominal operating charge from
the last column of Table R-14. This is the
maximum weight of liquid refrigerant which
can be allowed in the shell during the operat-
ing period covering some of the lower tubes.
Larger shell diameters or separate receivers
may be used where greater storage capacity is
needed during operation.

c. Determine the pump down capacity
from Table R-14. If less than the total weight
of refrigerant to be used in the system and
provision for complete pump-down are re-
quired, an additional receiver should be used.

Example 14-9. Select an R-12 condenser
to meet the following conditions:

Refrigeration load 30 tons

Condensing temperature 100° F

Suction temperature 30° F
Water available 2 gpm/ton

river water reasonably clean at 78°F

Maximum tube length 12 ft

Maximum water pressure drop 7.5 psi

Solution

1. From Fig. 2, the correction factor for
30° F suction temperature and 100° F con-
densing temperature is 1.013.

Corrected tons 30 x 1.013 = 30.4 tons

2. From Fig. 3, for 2 gpm/ton the water
temperature rise is found to be 14.4°.

Total gpm 30.4 x 2 = 60.8
Water "out" temperature 78 plus 14.4

= 92.4° F

3. GTD 100 - 78 = 22°
LTD 100 - 92.4 = 7.6°

From Table 11-1, METD = 13.55° F

4. Refer to Table R-14. Use of four passes
will usually give an economical selection for
75° F water in and 95° F water out which
approximates the required water conditions.
Note that a lOf shell will probably be needed.
This shell has sixty tubes.

Total gpm x number of passes

gpm per tube — — ^ — - — ; — : f

or r Number of tubes in condenser

_ 60.8 x 4

= 4.05 gpm per tube

5. Referring to Table 14-1, for clean river
water and over 3 fpm velocity, the suggested
scale factor is 0.001.

6. From Fig. 1, the V factor for 4.05 gpm per
tube and 0.001 scale factor is 121.5 Btu per hour
per square foot of extended surface per °F
METD.

7. Square feet required

Corrected tons x 14,400

U factor x METD

_ 30.4 x 14,400
~ 121.5 x 13.55

= 266 sq ft

8. Referring to Table R-14, a Model STF-
1010 has 289 sq ft external tube surface and
should be selected. When installed the water
connection should be made for four-pass
operation.

9. (a). For water pressure drop, refer to Fig.
4 and note that the pressure drop for 4.05 gpm
per tube in an STF-1010 condenser connected
for four passes is 7.1 psi. (b). Table R-14 shows
a nominal operating charge of 38 lb of R-12,
which will normally be sufficient for a 30-ton

installation. However, if more operating
storage is needed, a separate receiver may be
chosen, or alternately a different condenser
selection may be made if more economical, (c).
Table R-14 also shows pump-down capacity
which is 252 lb of R-12. Usually this will be
sufficient, but if greater pump-down capacity is
required, a separate receiver tank must be used.

14-13. Simplified Ratings. Simplified ratings,
based on the horsepower of the compressor
driver, are available for most air-cooled and
water-cooled condensers, particularly in smaller
sizes. Since the power required by the com-
pressor varies with both the evaporator load and
the compression ratio, it provides a reasonable
index of the condenser load at all operating
conditions. Table R-l 5, which applies to double-
tube condensers of the type shown in Fig. 14-12,
is a typical simplified condenser rating table.
14-14. Cooling Towers. Cooling towers are
essentially water conservation or recovery de-
vices. Warm water from the condenser is
pumped over the top of the cooling tower from
where it falls or is sprayed down to the tower
basin. The temperature of the water is reduced
as it gives up heat to the air circulating through
the tower.

Although there is some sensible heat transfer
from the water to the air, the cooling effect in a
cooling tower results almost entirely from the
evaporation of a portion of the water as the
water falls through the tower. The heat to
vaporize the portion of water that evaporates is
drawn from the remaining mass of the water so
that the temperature of the mass is reduced.
The vapor resulting from the evaporating pro-
cess is carried away by the air circulating
through the tower. Since both the temperature
and the moisture content of the air are increased
as the air passes through the tower, it is evident
that the effectiveness of the cooling tower
depends to a large degree on the wet bulb tem-
perature of the entering air. The lower the wet
bulb temperature of the entering air, the more
effective is the cooling tower.

The efficiency of a cooling tower is influenced
by all the factors governing the rate at which the
water will evaporate into the air (see Section
4-8). Some of the factors which determine
cooling tower efficiency are: (1) the mean
difference in vapor pressure between the air and
the water in the tower, (2) the amount of

CONDENSERS AND COOLING TOWERS 259

exposed water surface and the length (time) of
exposure, (3) the velocity of the air passing
through the tower, and (4) the direction of the
air flow with relation to the exposed water
surface (parallel, transverse, or counter).

For any given water temperature entering the
tower, the vapor pressure difference is essentially
a function of the wet bulb temperature of the
entering air. In general, the lower the entering
wet bulb temperature, the greater the vapor
pressure differential and the greater the tower
capacity.

The exposed water surface includes: (1) the
surface of the water in the tower basin, (2) all
wetted surfaces in the tower, and (3) the com-
bined surface of the water droplets falling
through the tower.

Theoretically, the lowest temperature to
which the water can be cooled in a cooling
tower is the wet bulb temperature of the entering
air, in which case the water vapor in the leaving
air will be saturated. In actual practice, it is not
possible to cool the water to the wet bulb tem-
perature of the air. In most cases, the tem-
perature of the water leaving the tower will be
7° to 10° F above the wet bulb temperature of
the entering air. Too, the air leaving the tower
will always be somewhat less than saturated.

The temperature difference between the tem-
perature of the water leaving the tower and the
wet bulb temperature of the entering air is
called the tower "approach." As a general rule,
all other conditions being equal, the greater the
quantity of water circulated over the tower the
closer the leaving water temperature approaches
the wet bulb temperature of the air. However,
the quantity of water which can be economically
circulated over the tower is somewhat limited by
the power requirements of the pump.

The temperature reduction experienced by the
water in passing through the tower (the differ-
ence between the entering and leaving water
temperatures) is called the "range" of the tower.
Naturally, to maintain equilibrium in the con-
denser water system, the tower "range" must
always be equal to the temperature rise of the
water in the condenser.*

The load on a cooling tower can be approxi-
mated by measuring the water flow rate over the

* Except where a condenser by-pass is used. See
Section 14-17.

260 PRINCIPLES OF REFRIGERATION

tower and the entering and leaving water tem-
peratures. The following equation is applied :

Tower load(Btu/min) = flow rate(gpm)
x 8.33 x (entering water temperature
— leaving water temperature) (14-12)

Example 14-10. Determine the approxi-
mate load on a cooling tower if the entering and
leaving water temperatures are 96° F and 88° F,
respectively, and the flow rate of the water over
the tower is 30 gpm.

Solution. Applying
14-12, the tower load
(Btu/min)

= 30 x 8.33

x (96 - 88)
= 2000 Btu/min

Since the load on the tower is equal to the
load on the condenser, the approximate refrig-
erating capacity of the system can be computed
by dividing the tower load by the condenser load
in Btu/min/ton corresponding to the operating
conditions of the system.

Example 14-1 1. Compute the refrigerating
capacity of an R-13 system operating on the
cooling tower of Example 14-10, if the evaporat-
ing and condensing temperatures are 20° F and
110° F, respectively.

Solution. From Fig. 14-1,
the load on the condenser

= 247 Btu/min/ton

The approximately refrig-
erating capacity of the
system

Tower load (Btu/min)

Condenser load (Btu/min/ton)
2000

247
= 8.1 tons

Since the heat absorbed per pound of water
evaporated is approximately 1000 Btu, assuming
a condenser load of 250 Btu/min/ton, the
quantity of water evaporated per ton of refrig-
eration (evaporator) is approximately 0.25 lb
per minute or 2 gal per hour.

In addition to the water lost by evaporation,
water is lost from the cooling tower by "drift"
and by "bleed-off." A small amount of water in
the form of small droplets is entrained and
carried away by the air passing through the
tower. Water lost in this manner is called the
drift loss. The amount of drift loss from a tower

depends on the design of the tower and the wind
velocity.

"Bleed-off" is the continuous or intermittent
wasting of a certain percentage of the circulated
water in order to avoid a build-up in the con-
centration of dissolved mineral solids and other
impurities in the condenser water. Without
bleed-off the concentration of dissolved mineral
solids in the condenser will build up quite rapidly
as a result of the evaporation taking place in the
cooling tower. Since the scaling rate is propor-
tional to the quality of the water, as the concen-
tration of mineral solids in the water increases
the scaling rate also increases.

The amount of bleed-off required to maintain
the concentration of dissolved mineral solids at
a reasonable level depends upon the cooling
range, the water flow rate, and the initial water
conditions. Suggested bleed-off rates for various
cooling ranges are given in Table 14-2. To
determine the quantity of water loss by bleed-
off, multiply the water flow rate over the tower
by the factor obtained from Table 14-2.

Example 14-12. Determine the quantity of
water lost by bleed-off if the water flow rate over
the tower is 30 gpm and the range is 10° F.

Solution. From
Table 14-1, the percent
bleed-off required = 0.33 %

The quantity of
water lost by bleed-off = 30 gpm x 0.0033

= 0.099 gpm

The bleed-off line should be located in the hot
water return line near the top of the tower so
that water is wasted only when the pump is
running (Fig. 14-8).

Make-up water, to replace that lost by evapo-
ration, drift, and bleed-off, is piped to the
tower basin through a float valve which tends to
maintain a constant water level in the basin.
14-15. Cooling Tower Design. According to
the method of air circulation, cooling towers are
classified as either natural draft or mechanical
draft. When air circulation through the tower
is by natural convection, the tower is called a
natural draft or atmospheric tower. When air
circulation through the tower is by action of a
fan or blower, the tower is called a mechanical
draft tower. Mechanical draft towers may be
further classified as either induced draft or
forced draft, depending on whether the fan or

Hot water in

Cold water out

Make-up water
from city main

Fig. 14-13. Natural draft-cooling tower.

blower draws the air through the tower or forces
(blows) it through. A schematic diagram of a
spray-type natural draft tower is shown in Fig.
14-13. Schematic diagrams of induced draft and
forced draft towers are shown in Figs. 14-14 and
14-15, respectively.

In the spray-type atmospheric tower, the
warm water from the condenser is pumped to
the top of the tower where it is sprayed down
through the tower through a series of spray
nozzles. Since the amount of exposed water
surface depends primarily on the spray pattern,
a good spray pattern is essential to high effi-
ciency. The type of spray pattern obtained
depends on the design of the nozzles. For most
nozzle designs, a water pressure drop of 7 to
101b per square inch will produce a suitable
spray pattern.

Some natural draft towers contain decking or
filling (usually of redwood) to increase the

CONDENSERS AND COOLING TOWERS 261

amount of wetted surface in the tower and to
break up the water into droplets and slow its
fall to the bottom of the tower. Atmospheric
towers containing decking are called "splash-
deck." Often, in splash-deck towers, no spray
nozzles are used and the water is broken up into
droplets by the "splash-impact" method.

The quantity and velocity of the air passing
through a natural draft cooling tower depend
on the wind velocity. Hence, the capacity of a
natural draft tower varies with the wind velocity,
as does the amount of "drift" experienced. Too,
natural draft towers must always be located out-
of-doors in places where the wind can blow
freely through the tower. In commercial appli-
cations, roof installations are common.

Since air circulation through mechanical
draft towers is by action of a fan or blower,
small mechanical draft towers can be installed
indoors as well as out-of-doors, provided that
an adequate amount of outside air is ducted into
and out of the indoor location. Too, since
larger air quantities and higher velocities can be
used, the capacity of a mechanical draft tower
per unit of physical size is considerably greater
than that of the natural draft tower. In addition,
most mechanical draft towers contain some sort
of decking or fill to improve further the effi-
ciency. Spray eliminators must be used in
mechanical draft towers to prevent excessive
drift losses.

14-16. Cooling Tower Rating and Selection.
Table R.-16 contains rating data for the spray-
type, natural draft cooling tower illustrated in
Fig. 14-13 and is a typical cooling tower rating
table. Notice that the tower ratings are given
in tons, based on a heat transfer capacity of 250

Water in

Water
distributor

Fig. 14-14. Small Induced
draft-cooling tower.

"Air out

262 PRINCIPLES OF REFRIGERATION

Air out

\ \ M f t f t t t f,

I Spray
>Teliminators

Wood fill

Fig. 14-15. Forced draft-
cooling tower.

Air in

Water out

Btu/min/ton. Nominal tower ratings are based
on a 3 mi per hour wind velocity, and 80° F
design wet bulb temperature, and a water flow
rate over the tower of 4 gpm per ton. Tower
performance at conditions other than those
listed in the table can be determined by using
the rating correction chart that accompanies the
table.

To select the proper tower from the rating
table, the following data must be known:

1 . Desired tower capacity in tons (compressor
capacity)

2. Design wet bulb temperature

3. Desired leaving water temperature (con-
denser entering water temperature or tower
approach)

1. Desired flow rate over the tower (gpm)

2. Design wet bulb temperature

3. Desired entering and leaving water tem-
peratures (tower cooling range and tower
approach)

Example 14-13. From Table R-16, select a
cooling tower to meet the following conditions:

1. Required tower capacity =20 tons

2. Design wet bulb

temperature = 78° F

3. Desired leaving water

temperature = 86° F

Solution. From Table R-16, select tower,
Model #CSA-66, which has a capacity of 20.7
tons at the desired conditions when the flow
rate over the tower is 3 gpm per ton. Hence, for
20-tons capacity, a total of 60 gpm (20 x 3)
must be circulated over the tower. As shown in
the table, the entering water temperature will be
approximately 96° F.

Exam pie 1 4- 1 4. It is desired to cool 90 gpm
from 96° F to 86° F when the design wet bulb is
78° F. Select the proper tower from Table
R-16.

Solution
Tower range

Tower approach

From rating correc-
tion chart, range-ap-
proach factor

From wet bulb cor-
rection chart, wet bulb
factor

Nominal gpm

From Table R-16, for
select tower, Model #SA-68

= 96 - 86 = 10°

= 86 - 78 = 8°

= 1.1

= 1.04

= 90 x 1.1 x 1.04

= 103 gpm

103 gpm nominal,

Example 14-15. It is required to cool water
for 30 tons at 5 gpm/ton to a 5° F approach of
an 80° F wet bulb. Select the proper tower from
Table R-16.

Solution

Total gpm required
for 30 tons at 5 gpm/
ton

From rating cor-
don chart, rating cor-
rection factor for 5
gpm/ton and 5° ap-
proach

From wet bulb
correction chart, wet
bulb correction factor

Nominal gpm

30 x 5 = 150 gpm

= 1.15

1.0

150 x 1.15 x 1.0

172.6 gpm

From Table R-16, for 172.6 gpm nominal,
select tower Model #SA-612.
14-17. Condenser By-Pass. For any given
tower range and approach, the entering and
leaving water temperatures will depend only on
the wet bulb temperature of the air. Hence, in
regions (particularly coastal areas) where the
outdoor wet bulb temperature is relatively high,
a closer approach to the wet bulb temperature is
required in order to maintain a reasonable con-
densing temperature with an economical con-
denser size than in areas where the wet-bulb
temperature is lower. It has already been shown
that, in general, the greater the quantity of water
circulated over the tower per unit of capacity
the closer the leaving water temperature will
approach the wet bulb temperature. Therefore,
in regions having a high wet bulb temperature,
it is usually desirable to circulate a greater
quantity of water over the tower than can be
economically circulated through the condenser
because of the excessive pumping head encoun-
tered. This can be accomplished by installing
a condenser by-pass line as shown in Fig. 14-8.
Through the use of a condenser by-pass, a
certain, predetermined portion of the water
circulated over the tower is permitted to by-pass
the condenser, thereby reducing the over-all
pumping head.

The advantage of the condenser by-pass is
that it makes possible the maintenance of
reasonable condensing temperatures with mod-
erate condenser and tower sizes without greatly
increasing the pumping head. The quantity of
water flowing through the by-pass is regulated
by the hand valve in the by-pass line. Once the
hand valve has been adjusted for the proper flow
rate through the by-pass, the handle should be
removed from the valve so that the valve adjust-
ment cannot be changed indiscriminately. An
excessive amount of water flowing through the
by-pass will not only tend to starve the con-
denser and raise the condensing pressure, but it
may also cause the pump motor to become
overloaded, thereby rendering the entire system
inoperative. The desired flow rate through the
by-pass is determined by subtracting the flow
rate through the condenser from the flow rate
over the tower. This will be illustrated presently.

Since the cooling tower capacity must of
necessity be equal to the condenser capacity at
the design conditions, it follows that:

CONDENSERS AND COOLING TOWERS 263

Tower gpm x tower range x 500

= condenser gpm x condenser rise x 500
Eliminating the constant,

Tower gpm x tower range

= condenser gpm x condenser rise (14-13)

Example 14-16. A compressor on a refrig-
erating system has a capacity of 25 tons. The
design wet bulb temperature is 80° F. The
desired condenser water entering temperature is
87° F and the desired temperature rise through
the condenser is 10° F. Select a cooling tower
from Table R-16 and determine:

1. The total gpm circulated over the tower

2. The temperature of the water entering the
tower

3. The tower cooling range

4. The temperature of the water leaving the
condenser

5. The gpm circulated through the condenser

6. The gpm circulated through the by-pass

Solution. From Table R-16, tower, Model
#SA-58 has a capacity of 25 tons at an 80° F
wet bulb temperature and a 7° approach. This
capacity is based on a water flow rate of 4
gpm/ton and on a cooling range of 7.5°
(94.5 - 87).

Total gpm over the tower
for 25 tons

= 25 tons x 4 gpm/ton
= 100 gpm

From Table R-16, the
tower entering water tem-
perature

= 94.5° F
Tower range

= 94.5 - 87 - 7.5°
Water temperature leaving
condenser

= 87 + 10 = 97° F

Rearranging and applying
Equation 14-13, condenser
gpm

_ Tower gpm x tower range

Condenser rise
_ 100 x 7.5

10
= 75 gpm

Gpm circulated through
by-pass

= Tower gpm — condenser gpm
= 100-75
= 25 gpm

364 PRINCIPLES OF REFRIGERATION

Refrigerant
vapor in

Refrigerant

liquid Out

\^J|/ / Eliminators

Spray
i ,, I >: ■

Condensing coil

Air in

M-thL'-ut:

•.'iL.'.'J

!;-~.-I Water tank -_-_-_-_-:

?■•.. m Ti

Fig. 14-14. Schematic diagram of evaporative con-
denser.

14-18. Evaporative Condeneers. An evapo-
rative condenser is essentially a water Conser-
vation device and is, in effect, a condenser and
a cooling tower combined into a single unit. A
diagram of a typical evaporative condenser is
shown in Fig. 14-16,

As previously stated, both air and water are
employed in the evaporative condenser. The
water, pumped from the sump up to the spray
header, sprays down over the refrigerant coils
and returns to the sump. The air is drawn in
From the outside at the bottom of the condenser
by action of the blower and is discharged back
to the outside at the top of the condenser. In
some cases, both pump and biower are driven
by the same motor. In others, separate motors
are used. The eliminators installed in the air
Stream above the spray header arc to prevent
entrained water from being carried over into the
blower. Art alternate arrangement, with the
blower iocated on the entering air side oF the
condenser, is shown in Fig. 14-17.

Although the actual thermodynamic pro-
cesses taking place in the evaporative condenser
are somewhat complex., the fundamental process
is that of evaporative cooling. Water is evapo-
rated from the spray and from the wetted surface

of the condenser into fhe air, the source of the
vaporising heat being the condensing refrigerant
in the condenser coil.

The cooling produced is approximately 1000
titu per pound of water evaporated. All the heat
given up by the refrigerant in the condenser
eventually leaves the condenser as either sensible
heat or latent heat (moisture) in the discharge
air. Since both the temperature and the mois-
ture content of the air are increased as the air
passes through the condenser, the effectiveness
of the condenser depends, in part, on the wet
bulb temperature of the entering air. The lower
the wet bulb temperature of the entering air the
more effective is the evaporative condenser.

To facilitate cleaning and scale removal, the
condensing coil is usually made up of bare
rather than fumed tubing. The amount of coil
surface used per ton of capacity varies with the
manufacturer and depends to a large extent on
the amount of air and water circulated.

Generally, the capacity of the evaporative
condenser increases as the quantity of air cir-
culated through the condenser increases. As a
practical matter, the maximum quantity of air

Fig. 14-17. Cutaway view of "Dri-Fan" evaporative

condenser. Funnel-shaped overflow drain provide]
au (ornate bleed-off. (Courtesy/ Refrigeration Engln-

Bering, Inc. A proprietary design of Refrigeration
Engineering, Inc.)

CONDENSERS AND COOLING TOWERS 265

which can be circulated through the condenser
is limited by the horsepower requirements of the
fan and by the maximum air velocity that can be
permitted through the eliminators without the
carry over of water particles.

The quantity of water circulated over the
condenser should be sufficient to keep the tube
surface thoroughly wetted in order to obtain
maximum efficiency from the tube surface and to
minimize the rate of scale formation. However,
a water flow rate in excess of the amount
required for adequate wetting of the tubes will
only increase the power requirements of the
pump without materially increasing the con-
denser capacity.

Assuming a condenser load of 15,000 Btu per
hour per ton, the water lost by evaporation is
approximately 15 lb (2 gal) per hour per ton
(15,000/1000). In addition to the water lost by
evaporation, a certain amount of water is lost by
drift and by bleed-off. The amount of water
lost by drift and by bleed-off is approximately
l.S to 2.5 gal per hour per ton, depending upon
the design of the condenser and the quality of
water used. Hence, total water consumption
for an evaporative condenser is between 3 and 4
gal per hour per ton.

Some evaporative condensers are available
equipped with desuperheating coils, which are
usually installed in the leaving air stream. The
hot gas leaving the compressor passes first
through the desuperheating coils where its tem-
perature is reduced before it enters the con-
densing coils. The desuperheating coils tend to
increase the over-all capacity of the condenser
and reduce the scaling rate by lowering the
temperature of the wetted tubes. Too, often the
receiver tank is located in the sump of the
evaporative condenser in order to increase the
amount of liquid subcooling.
14-19. Rating and Selection of Evaporative
Condensers. Table.R-17 is a typical evapo-
rative condenser rating table. Notice that the
ratings are based on the temperature difference
between the condensing temperature and the
design wet bulb temperature. The following
sample selection is reprinted directly from the
manufacturer's catalog data:*

Example 14-17. Select an evaporative con-
denser for the following conditions:

* McQuay Products.

6-ton evaporator load (Refrigerant- 12)
20° evaporator temperature
78° entering wet bulb temperature
105° F condensing temperature

Solution. Since the rating table is in terms of
evaporator load at 40° F, it is necessary to
correct for other evaporator temperatures by
using a correction factor from R-17B as follows:
Tons x evaporator correction factor

•= Rating table tons
Therefore, 6 x 1.05 = 6.3 tons.

Referring to Table R-17A, the E-135F has a
capacity of only 5.6 tons at 78° F entering wet
bulb and 105° F condensing temperature. It
does, however, have the required capacity of 6.3
tons at between 105° F and 110° F condensing
temperature.

The compressor ratings should then be
checked to see if the compressor originally
selected has the required capacity at between
105° F and 110° F condensing temperature. If
not, it will be necessary to select the next larger
size evaporative condenser or compressor to do
the job.

The next larger size evaporative condenser,
the E-270F, has a capacity of 11.2 tons at the
given conditions; however, the required capa-
city of 6.3 tons will be obtained at a condensing
temperature between 90 and 95° F. The com-
pressor selection should then be made for these
conditions.

14-20. Water Regulating Valves. The water
flow rate through a water-cooled condenser on a
waste water system is automatically controlled
by a water regulating valve (Fig. 14-18). The
valve is installed on the water line at the inlet of
the condenser and is actuated by the compressor
discharge (Fig. 14-7). When the compressor
is in operation, the valve acts to modulate the
flow of water through the condenser in response
to changes in the condensing pressure. An
increase in the condensing pressure tends to
collapse the bellows further and open the valve
wider against the tension of the range spring,
thereby increasing the water flow rate through
the condenser. Likewise, as the condensing
pressure decreases, the valve moves toward the
closed position so that the flow rate through the
condenser is reduced accordingly. Although
the regulating valve tends to maintain the con-
densing pressure constant within reasonable
limits, the condensing pressure will usually be
considerably higher during periods of peak
loading than during those of light loading.

Hi PRINCIPLES OF REFRIGERATION

Flf. 14-lfl. Typical threaded-type water regulating valve. Larger sixes are available with, flange connection*.
{a) Cross-sectional view showing principal parts, (b) Exterior view, (Courtesy Penn Contrail, Inc.)-

When the compressor cycles off, the water
valve remains open and water continues la flow
through ihc condenser until the pressure in the
condenser is reduced to a certain predetermined
minimum, at which time the valve closes off
completely and shuts o FT the water flow. When
the compressor cycles on again, ihc water valve
remains closed until the pressure in the con-
denser builds up to the valve opening pressure,
at which time the valve opens and permits water
to flow Lh rough the condenser. The opening
pressure of the valve is approximately 7 psi
above the shut-oil pressure.

The water valve is set for the desired shut-off
pressure by adjusting the tension of the range
spring. The minimum operating pressure for
the valve, that is, the shut-off pressure, must be
set high enough so that the valve will not remain
open and permit water to flow through the con-
denser when the compressor is on the off cycle.
Since the saturation temperature of the refrig-

erant in the condenser can never be lower than
the ambient temperature at the condenser, the
shut-off point of the water valve should be set at
a saturation pressure corresponding to the maxi-
mum ambient temperature in the summertime
at the condenser location. Too, the shut-off
pressure of the valve must be high enough so
that the minimum condensing temperature in
the wintertime is sufficiently high to provide a
pressure differential across the refrigerant con-
trol large enough to assure its proper operation.

The capacity of water regu Sating valves varies
with the size of the valve and the pressure drop
across the valve orifice. The available pressure
drop across the valve orifice is determined by
subtracting the pressure drop through the con-
denser and water piping from the total pressure
drop available at the water main.

Water regulating valves are usually selected
from flow charts (Table fl-lS), In order to
select the proper valve from the flow chart, the

CONDENSERS AND COOLING TOWERS 267

following data must be known: (1) the desired
water quantity in gpm; (2) the maximum
ambient temperature in the summertime; (3)
the desired condensing temperature; and (4)
the available water pressure drop across the
valve.

The following selection procedure and sample
selection are reprinted directly from the litera-
ture of the manufacturer:*

1. Draw horizontal line across upper half of
Flow Chart (Table R-18) through the required
flow rate.

2. Determine refrigerant condensing pressure
rise above valve opening point.

a. Valve closing point (to assure closure
under all conditions) must be the refrigerant
condensing pressure equivalent to the highest
ambient air temperature expected at time of
maximum load. Read this in psig from "Satu-
rated Vapor Table" for refrigerant selected.

b. Read from the same table the operating
condensing pressure corresponding to selected
condensing temperature.

c. Valve opening point will be about 7 psi
above closing point.

d. Subtract opening pressure from operat-
ing pressure. This gives the condensing
pressure rise.

3. Draw horizontal line across lower half of
Flow Chart through this value.

4. Determine the water pressure drop through
the valve — this is the pressure actually available
to force the water through the valve.

a. Determine the minimum water pressure
available from city mains or other source.

b. From condensing unit manufacturer's
tables read pressure drop through condenser
corresponding to required flow.

c. Add to this estimated or calculated drop
through piping, etc., between water valve and
condenser, and from condenser to drain (or
sump of cooling tower).

d. Subtract total condenser and piping drop
from available water pressure. This is the
available pressure drop through the valve.

Example. 14-18. The required flow for an
R-12 system is found to be 27 gpm. Condensing
pressure is 125 psig and the maximum ambient
temperature estimated at 86° F. City water

» By permission of Penn Controls, Inc., Goshen,
Indiana.

pressure is 40 psig and manufacturer's table
gives drop through condenser and accompany-
ing piping and valves as 15 psi. Drop through
installed piping approximately 4 psi. Select
proper size of water regulating valve from Table
R-18.

Solution

1. Draw a line through 27 gpm — see dotted
line, upper half of Flow Chart (Table R-18).

2. Closing point of valve is pressure of R-12
corresponding to 86° F ambient = 93 psig.

3. Opening point of valve is 93 + 7 = 100
psig.

4. Condensing pressure rise = 125 — 100 =
25 psi.

5. Draw line through 25 psi—see dotted line,
lower half of Flow Chart.

6. Available water pressure drop through
valve = 40 - 19 = 21 psi.

7. Interpolate just over the 20 psi curve —
circle on lower half of Flow Chart.

8. Draw vertical line upward from this point
to flow line — circle on Flow Chart marks this
intersection.

9. This intersection falls between curves for
1 in. and \\ in. valves. The 1J in. valve is re-
quired.

14-21. Condenser Controls. For reasons of
economy, the condensing medium is circulated
through the condenser only when the com-
pressor is operating. Hence, common practice
is to cycle the condenser fan and/or pump on
and off with the compressor. This is usually
accomplished by electrically interlocking the
fan and/or pump circuit with the compressor
driver circuit. Method of interlocking electrical
circuits are discussed in Chapter 21.

Whereas high pressure controls are always
desirable as safety devices on any type of system,
they are absolutely essential on all equipment
employing water as the condensing medium in
order to protect the equipment against damage
from high condensing pressures and tempera-
tures in the event that the water supply becomes
restricted or is shut-off completely. The high
pressure control has already been discussed in
Section 13-13.

If a refrigerating system is to function pro-
perly and efficiently, the condensing temperature
must be maintained within certain limits. As
previously described, high condensing tempera-
tures cause losses in compressor capacity and
efficiency, excessive power consumption, and,

268 PRINCIPLES OF REFRIGERATION

in some cases, overloading of the compressor
driver and/or serious damage to the compressor
itself.

An abnormally low condensing temperature,
on the other hand, will cause an insufficient
pressure differential across the refrigerant con-
trol (condensing pressure to vaporizing pres-
sure), which reduces the capacity of the control
and results in starving of the evaporator and
general unbalancing of the system.

As a general rule, low condensing tempera-
tures result from either one or both of two
principal causes: (1) low ambient temperatures
and (2) light refrigerating loads. Naturally, the
problem of low condensing temperatures is more
acute in the wintertime when the ambient
temperature and the refrigerating load are both
apt to be low.

To maintain the condensing temperature at a
sufficiently high level, it is necessary to make
some provision for reducing or controlling the

Air-cooled
condenser

From

condenser

outlet

capacity of the condenser during periods when
the ambient temperature is low and/or the refrig-
erating load is light. Although the methods
employed to control the capacity of the con-
denser vary somewhat with the type of con-
denser used, all involve reducing either the
quantity of condensing medium circulated or
the amount of effective condensing surface.
Condenser capacity control devices are usually
actuated by pressure or temperature controls
which respond to condensing pressure or
temperature.

With regard to air-cooled condensers, the
condensing temperature is maintained within
the desired limits by varying the air quantity
through the condenser or by causing a portion
of the condenser to become filled with liquid so
as to reduce the amount of effective condensing
surface.

The air quantity through the condenser is
varied by cycling the fan or blower or by the use

Modulating

control valve

(open on drop

in pressure)

Fig. 14-19. Winterstat control
of air-cooled condensers, (a)
Loop Winterstat may be used
wherever 3 feet of head room
is available above the top of the
condenser. This type is the
simplest and lowest in cost.
(b) No-loop Winterstat is
employed where head room
is not available above con-
denser. Valves and W are
supplied as an integral unit and
must be mounted at the level
of the liquid outlet of the
condenser. (The Winterstat is
a proprietary design of the
Kramer Trenton Company and
is manufactured under the
following patent numbers:
2,564,310; 2,761,287; and
2,869,330.)

Constant
inlet pressure
throttling valve

To receiver

(b)

CONDENSERS AND COOLING TOWERS 269

Fig. 14-20. Pressure stabilizer.
(A proprietary design of
Dunham-Bush, Inc.) (Courtesy
Dunham-Bush, Inc.)

of dampers placed in the air stream. Because it
tends to cause large fluctuations in the con-
densing temperature, cycling of the fan cannot
be recommended as a means of controlling the
capacity of air-cooled condensers. Modulating
dampers installed in the air stream provide
satisfactory control of the air quantity in many
cases, although some difficulty is experienced
with dampers when the condenser is exposed to
high wind velocities.

A more satisfactory method of controlling
the capacity of air-cooled condensers is to vary
the amount of effective condensing surface by
causing the liquid refrigerant to back up into
the lower portion of the condenser whenever the
condensing pressure drops below the desired
minimum. To accomplish this, one design of
capacity control employs a modulating valve
installed in a by-pass line between the inlet and
outlet of the condenser (Fig. 14-19). As the
receiver pressure falls, the modulating valve
opens and allows high-pressure vapor from the
compressor discharge to flow through the by-
pass line, thereby restricting the flow of liquid
refrigerant from the condenser and causing the
liquid to back up into the lower portion of the
unit. The amount of discharge vapor by-passed,
and therefore the amount of liquid refrigerant
retained in the lower portion of the condenser,
is automatically controlled by the modulating
valve and depends upon the receiver tank
pressure.

Another device used to restrict the amount of
effective condensing surface is called a "pressure
stabilizer" (Fig. 14-20). The following descrip-
tion of the operation of the pressure stabilizer
is reprinted directly from the manufacturer's
engineering data.*

The pressure stabilizer is a heat transfer
surface which transfers the heat from the hot gas
discharge of the compressor to the subcooled
liquid leaving the condenser. This heat exchange

* Courtesy Dunham-Bush, Inc.

is controlled by the regulating valve installed
between the condenser and the receiver. This
valve is set at the desired operating pressure, and
throttles from the open position to the closed
position as the condensing pressure drops. The
throttling action backs up the liquid in the con-
denser, thus reducing the amount of effective
condensing surface. The subcooled liquid
coming from the condenser is forced through
the heat exchanger portion of the pressure
stabilizer and receives enough heat from the hot
gas to satisfactorily establish the balanced
pressure temperature relationship in the receiver.
This assures satisfactory condensing pressure
and a solid column of liquid at the refrigerant
control.

The pressure stabilizer is designed with a pre-
determined pressure drop to insure against
liquid refrigerant reheating during warm weather
operations. During high ambient air tempera-
tures, where the condensing temperature is above
the setting of the regulating valve, the liquid
flows through the valve, which is fully open, and
thereby by-passes the heat exchanger section
(Fig. 14-21a). In Fig. 14-216, as the ambient
temperature drops to 50° F the condensing tem-
perature drops below the setting of the regulat-
ing valve. The valve then modulates toward
the closed position, and this action limits the
flow of liquid through the regulating valve.
Consequently, the liquid backs up in the con-
denser until the condensing surface is reduced
approximately 60%. The liquid which is forced
to pass through the heat exchanger section is
then heated up to the saturation temperature.

When the ambient temperature drops to 0° F
(Fig. 14-2 lc), the regulating valve throttles to
hold 120 psi in the condenser. The liquid logs
in the condenser so that approximately 10% of
the surface is utilized to condense the hot gas.

With regard to evaporative condensers, capa-
city control is best obtained through regulation
of the air quantity through the condenser, which
can be accomplished either by cycling the blower

270 PRINCIPLES OF REFRIGERATION

90° Amb., R-12
110* Cond. temp.

I r' '■ '■"■ i -' vr ■■■

*' -■;'■■ 'v )

136 g ^ ' ■'

PTTX. ........ ....'.j

50* Amb., R-12
102* Cond. temp.

[ ■;■ ■ •■—

l..."..'.'..-"l^J,»l

,II'I.H'..I| ^T"f J

0* Amb., R-12
102* Cond. temp.

- ffi'A'.",;.',..',-.'.'M

Fig. 14-21. Air-cooled condenser control employing
pressure stabilizer. (Courtesy Dunham-Bush, Inc.)

or by installing dampers in the air stream. Of
the two methods, the latter is usually the most
satisfactory, especially where modulating dam-
pers are used and the air quantity can be varied
through a wide range.

Cycling of the pump as a means of controlling
the capacity of an evaporative condenser cannot
be recommended. Each time the pump cycles
off a thin film of scale is formed on the con-
denser tubes. Consequently, frequent cycling of
the condenser pump greatly increases the scaling
rate, which reduces the efficiency of the con-
denser and greatly increases maintenance costs.
With reference to water-cooled condensers,
recall that for a given load and condensing
surface, the condensing temperature varies with
the quantity and temperature of the water
entering the condenser. Where waste water is
used, the modulating action of the water-
regulating valve controls the water flow rate
through the condenser and maintains the con-
densing temperature above the desired minimum
so that low condensing temperatures are not
usually a problem with waste water systems. On
the other hand, since the flow rate of the water
through the condenser on a recirculating water
system is maintained constant, the condensing
temperature decreases as the temperature of the
water leaving the tower decreases. Therefore,
when the ambient air temperature is low, the
condensing temperature will also be low unless
some means is provided for restricting the flow
rate through the condenser or for increasing the
temperature of the water leaving the tower.

One method of controlling the condensing
temperature in a recirculating water system is to
install a water-regulating valve in the water line
at the inlet to the condenser. The modulating
action of the water valve will restrict the water
flow rate through the condenser in response to a
drop in the condensing pressure. When a water-
regulating valve is used in a recirculating water
system, the pressure drop through the valve
must be taken into account in computing the
total pumping head.*

Where mechanical draft cooling towers are
used, the condensing temperature can be main-
tained at the desired level through regulation of
the tower leaving water temperature. As in the
case of the evaporative condenser, this can be
accomplished by cycling the tower fan or by
installing dampers in the air stream.

* Except in those cases where they have a specific
function, water-regulating valves should never be
used in recirculating water systems, since they tend
to restrict the water flow and increase the pumping
head unnecessarily.

CONDENSERS AND COOLING TOWERS 271

14-22. Winter Operation. When the com-
pressor and/or condenser are so located that they
are exposed to low ambient temperatures, the
pressure in these parts may fall considerably
below that in the evaporator during the com-
pressor off-cycle. In such cases, the liquid
refrigerant, which otherwise would remain in
the evaporator, very often tends to migrate to
the area of lower pressure in the compressor and
condenser. With no liquid refrigerant in the
evaporator, an increase in evaporator tempera-
ture is not reflected by a corresponding increase
in the evaporator pressure, and, where the sys-
tem is controlled by a low pressure motor
control, the rise in evaporator pressure may not
be sufficient to actuate the control and cycle the
system on in response to an increase in the
evaporator temperature.

Corrective measures are several. One is to
install a thermostatic motor control in series
with the low pressure control. The thermostat
is adjusted to cycle the system on and off,
whereas the low pressure control serves only as a
safety device. Another, and usually more prac-
tical, solution is to isolate the condenser during
the off-cycle. One method of isolating the con-
denser during the off-cycle is illustrated in Fig.
14-22. The check valve (Q in the condenser
liquid line prevents the refrigerant from boiling
off in the receiver and backflowing to the con-

Air-cooled
condenser

Modulating valve

(open on rise of

inlet pressure)

ggB%g

Check valve

To liquid_
receiver

Modulating by-pass

valve-open of drop

in outlet pressure

From _y
discharge

Fig. 14-22. Sure-start WintersUt provides normal
head and receiver pressures when the compressor
starts by allowing the compressor to impose its full
discharge pressure on the liquid through the open
(W) valve. When the receiver pressure is up to
normal, the (R) valve opens and allows the discharge
gas to flow to the condenser. (Courtesy Kramer
Trenton Company.)

^Modulating
dampers

Water bleeds

Drain-

Float valve

Overflow

rm\

Sump tank

•k

Pump

Fig. 14-23. Evaporative condenser equipped with
modulating dampers for capacity control. Protected
auxiliary pump is designed to prevent freezing during
winter operation. (Courtesy Refrigeration Engineer-
ing Inc.)

denser during the off-cycle. The (R) valve,
which closes on drop of pressure at the valve
inlet, closes when the compressor stops, pre-
venting the flow of refrigerant from the evapo-
rator, through the compressor valves and
discharge line, into the condenser. With the
condenser isolated, the evaporator pressure can
build up and start the compressor regardless of
the ambient temperature at the condenser.

Another and rather obvious problem con-
cerning the operation of evaporative condensers
and cooling towers in the wintertime is the
danger of freezing when the equipment is
exposed to freezing temperatures. In general,
the measures employed to prevent freezing are
similar to those used to prevent low condensing
temperatures, that is, controlling the air quantity
through the tower by the use of dampers or by
cycling the fan. In addition, an auxiliary sump
must be installed in a warm location and the
piping arranged so that the water drains by
gravity into the auxiliary sump and does not
remain in the tower or condenser sump (Figs.
14-23 and 14-24).

14-23. Condenser and Tower Maintenance.
As a general rule, air-cooled condensers require
little maintenance other than regular lubrication
of the fan and motor bearings. However, the

272 PRINCIPLES OF REFRIGERATION

j ^ ** ^^^^^^™

Tower static
head

Additional
static head

Indoor
tank

Fig. 14-24. Protected indoor tank.

fan blades and condensing surface should be
inspected occasionally for the accumulation of
dust and other foreign materials. These parts
should be kept clean in order to obtain high
efficiency from the condenser.

Any type of condenser employing water is
subject to scaling of the condenser tubes, corro-
sion, and the growth of algae and bacterial
slime on all wetted surfaces. The latter is con-
trolled by frequent cleaning of the infected parts
and by the use of various algaecides which are
available commercially.

As previously stated, the scaling rate depends
primarily upon the condensing temperature and
the quality of water used. The scaling rate will
be relatively low where the condenser leaving
water temperature is below 100° F. Too, the
importance of providing for the recommended
amount of bleed-off cannot be overemphasized
with regard to keeping the scaling rate at a
minimum. In addition, a number of chemical
companies have products which when added to
the sump water considerably reduce the scaling
rate.

Scale can be removed from the condenser

tubes by applying an approved inhibited acid
compound, many of which are available in
either liquid or powder form. After the tower
or condenser sump has been drained, cleaned,
and filled with fresh water, the cleaning com-
pound can be added directly to the sump water.
The pump is then started and the cleaner is
circulated through the system until the system is
clean, at which time the sump is again drained,
flushed, and filled with clean water before the
system is placed in normal operation.

It should be pointed out that descaling com-
pounds have an acid base and should not be
allowed to contact grass, shrubs, or painted
surfaces. Therefore, it is usually advisable to
remove the cooling tower spray nozzles, if any,
in order to minimize the danger of damaging
shrubs or painted surfaces with drift from the
tower.

When rapid descaling of the condenser tubes
is required, an inhibited solution (18%) of
muriatic acid may be used. However, muriatic
acid should be used only on the condenser tubes.
The system pump should not be used to circu-
late the acid. A small pump having an acid
resistant impeller (brass or nylon) may be used
for this purpose (see Fig. 14-25). After the
condenser is clean, it should be flushed with
clean water or with an acid neutralizer as
recommended by the manufacturer.

Corrosion is usually greatest in areas near
salt water or in industrial areas where relative

Condenser -

Fig. 14-25. Apparatus for descaling condenser.

large concentrations of sulfur and other indus-
trial fumes are found in the atmosphere. Corro-
sion damage is minimized by regular cleaning
and painting of the affected parts and by
application of protective coatings of various
types.

PROBLEMS

1. An R-12 system is operating at an evaporator
temperature of 0° F and a condensing tempera-
ture of 100° F. From Chart 14-1, determine the
heat load on the condenser in Btu per minute
per ton of refrigeration. Ans. 257 Btu/min/ton

2. An R-22 system operating with a 40° F
evaporator and a 110° F condenser has an
evaporator load of 10 tons. Determine the heat
load on the condenser in Btu/hr.

Ans. 141,000 Btu/hr

3. The heat rejected to a water-cooled con-
denser is 120,000 Btu/hr. How many square
feet of effective tube surface must this condenser
have if the U factor of the condenser is 100
Btu/hr/sq ft/° F and the METD is 5° F at the
desired gpm? Ans. 240 sq ft

4. The heat load on the evaporator of an air
conditioning system is 60,000 Btu/hr. If the
coefficient of performance of the system is 4 : 1,
what is the heat load on the condenser in Btu/hr ?

Ans. 75,000 Btu/hr

5. An R-12 waste water system operating at a
40° F suction temperature and a 105° F con-
densing temperature has an evaporator load of
5 tons. If the condenser is selected for a 12° F
water temperature rise, how many gpm must be
circulated through the condenser?

Ans. 11.5 gpm

6. Seventy-two gallons of water per minute are
circulated through a water-cooled condenser.
If the temperature rise of the water in the con-
denser is 14° F, what is the heat load on the
condenser? Ans. 504,000 Btu/hr

7. An R-12 air conditioning system operating
with an evaporator temperature of 40° F and a
condensing temperature of 120° F has an
evaporator load of 60,000 Btu/hr. 4500 cfm of
air are circulated over the condenser. If the
temperature of the air entering the condenser is
90° F, compute: (a) the leaving air temperature
and (b) the METD.

Ans. (a) 104.6° F (b) 21.89° F

8. If the air-cooled condenser in Problem 7 has
a free face area of 5.5 sq ft, what is the velocity
of the air through the condenser?

Ans. 818 fpm

CONDENSERS AND COOLING TOWERS 273

9. From Table R-12, select an air-cooled con-
denser for a compressor having a capacity of
42,000 Btu/hr if the design suction and discharge
temperatures are 40° F and 130° F, respectively,
and the outdoor design dry bulb temperature
for the region is 95° F.

10. Select a shell-and-tube water-cooled con-
denser for an R-12 system to meet the following
conditions :

Refrigeration load and eva-
porator 60 tons
Evaporator temperature 40° F
Condensing temperature 1 10° F
Water quantity 2.5 gpm/ton
Untreated cooling tower water enters con-
denser at 85° F.

11. Rework Problem 10 using a condensing
temperature of 120° F.

12. A cooling tower and a water-cooled con-
denser (with by-pass) are operating with a
condenser load of 240,000 Btu/hr. Forty-eight
gpm are circulated through the condenser and
32 gpm are by-passed. The ambient wet bulb
temperature is 78° F and the tower approach is
7° F. Determine:

(a) The temperature of the water entering the
condenser. Ans. 85° F

(6) The temperature of the water leaving the
condenser. Ans. 95° F

(</) The tower range. Ans. 6° F

13. A compressor on a Refrigerant- 12 system
has a capacity of 50 tons. The design wet bulb
temperature is 78° F. The desired condenser
water entering temperature is 85° F and the
desired temperature rise through the condenser
is 12° F. Select a cooling tower from Table R-15
and determine:

(a) The total gpm circulated over the tower

(b) The temperature of the water entering the
tower

(d) The tower range

(e) The gpm circulated through the condenser
(/) The gpm by-passed

14. Select an evaporative condenser for the
following conditions:

Refrigerant- 12 system

Evaporator load — 10 tons

Evaporator temperature — 40° F

Wet bulb temperature of entering air — 78° F

Condensing temperature — 105° F

Fluid Flow,
Centrifugal Liquid
Pumps, Water and
Brine Piping

15-1. Fluid Pressure. The total pressure
exerted by any fluid is the sum of the static and
velocity pressures of the fluid, viz:

Pt =P* +Pv (15-1)

where p t = the total pressure
p, = the static pressure
p v = the velocity pressure

All flowing fluids possess kinetic energy and
therefore exert a force or pressure in the direc-
tion of flow. The pressure exerted by a fluid
which is the direct result of fluid motion or velo-
city is called the velocity pressure of the fluid.
Any pressure exerted by a fluid which is not the
direct result of fluid motion or velocity, regard-
less of the force causing the pressure, is called
the static pressure of the fluid. For fluids at rest
(static), the velocity pressure is equal to zero
and the total pressure is equal to the static
pressure. Whereas velocity pressure acts only
in the direction of flow, static pressure acts
equally in all directions. This is easily demon-
strated through the use of an example employing
a gravitational column.

It was shown in Chapter 1 that the action of
gravity on any body causes the body to exert
a force which is commonly referred to as the
weight of the body. For a solid material,
because of the rigid molecular structure, the

gravitational force or pressure is exerted in a
downward direction only. However, because of
the loose molecular structure of fluids, the gravi-
tation force or pressure exerted at any point in a
body of fluid acts equally in all directions — up,
down, and sideways, and always at right angles
to any containing surfaces. When no force
other than the force of gravity is acting on the
fluid, the pressure at any depth in a body of fluid
is proportional to the weight of fluid above that
depth. When an external force in addition to
the force of gravity is applied to the liquid, the
pressure at any depth in the fluid is proportional
to the weight of the fluid above that depth, plus
the pressure caused by the external force.

For example, assume that a flat-bottomed
container 1 sq ft in cross section and 10 ft high
is filled to the top with water at a temperature of
60° F (Fig. 15-1). Since water at 60° F has a
density of 62.4 lb per cubic foot, if the pressure
of the atmosphere on the surface of the water is
neglected, the total force acting on the bottom
of the tank due to the weight of the water alone
is 624 lb (10 x 62.4). Since the base area of the
tank is 1 sq ft, the pressure exerted on the
bottom of the tank is 624 psf or 4.33 psi
(624/144). Since this pressure acts equally in all
directions, it is exerted on the sides of the tank
at the base as well as on the bottom of the tank.

Assume now that level A in the water column
is exactly 1 ft below the surface of the water.
The volume and weight of water above this level
are 1 cu ft and 62.4 lb, respectively. Since this
weight of water is also evenly distributed over an
area of 1 sq ft, the fluid pressure acting in all
directions from any point at level A is 62.4 psf
or 0.433 psi. Similarly, the volume and weight
of water above level B, which is located 5 ft
below the surface of the water, are 5 cu ft and
3121b (5 x 62.4), respectively, and the fluid
pressure at this level is 312 fsf or 2.165 psi.

If the force exerted on the top of the water by
the pressure of the atmosphere is taken into
account, the pressure of the water at any level
in the tank will be increased by an amount equal
to the pressure of the atmosphere. Assuming
normal sea level pressure, the fluid pressures at
levels A and B are 15.129 psi (0.433 + 14.696)
and 16.861 (2.165 + 14.696), respectively, while
the pressure at the base of the tank is 19.026 psi
(4.33 + 14.696). However, it should be recog-
nized that since the pressure of the atmosphere

274

FLUID FLOW, CENTRIFUGAL LIQUID PUMPS, WATER AND BRINE PIPING 275

is exerted also on the outside of the tank the
pressure tending to burst the tank is still only
that resulting from the gravitational effect on
the water alone.

For any noncompressible fluid (liquid), the
pressure exerted by the fluid at any level in a
fluid column is directly proportional to the
depth of the fluid at that level.* Hence, the
pressure of a liquid at any level in a column of
liquid can be determined by multiplying the
depth at that level times the density of the fluid,
viz:

Pressure (psf) = depth (ft) x density (lb/cu ft)

(15-2)

Pressure (psi)

depth (ft) x density (lb/cu ft)
144

(15-3)

15-2. Head-Pressure Relationship. The

vertical distance between any two levels in a
column of liquid is called the "head" of the
liquid at the lower level with respect to the upper
level. For example, with respect to level B in
Fig. 1 5- 1 , the head of the water at the base of the
column is 5 ft. With respect to the top of the
column, the head of the water at the base of
the column is 10 ft. Similarly, with respect to the
top, the water heads at levels A and £ are 1 ft
and 5 ft, respectively.

Since the depth of the liquid at any level in a
liquid column is equal to the head of the liquid
at that level with respect to the top of the column,
the head can be substituted for depth in
Equation 15-3 and the following relationship
between head and pressure is established:

Pressure (psi) =

Head (ft) x density (3/cu ft)
144

(15-4)

„ T ,. ,, . Pressure (psi) x 144
Head (ft) = — — . *, /, — — - (15-5)
Density (lb/cu ft) K '

It is evident from the foregoing that there is a
definite and fixed relationship between the head
and the pressure of any liquid, the head-pressure
ratio for any given liquid being dependent upon
the density of the liquid. For example, in the
case of water, the head-pressure ratio is 2.31 ft

* This is not true of a compressible fluid because
the density of a compressible fluid varies with the
depth.

P =

Fig. 15-1. Illustrating head-pressure relationship.

to 1 psi. For mercury, the head-pressure ratio
is 2.04 in. to 1 psi. This means that a pressure
of 1 psi is equivalent, to head of 2.31 ft of water
column or 2.04 in. of mercury column. Con-
versely, a 1 ft column of water (1 ft water head)
is equivalent to 0.433 psi, whereas a 1 ft column
of mercury (1 ft mercury head) is equivalent to
24.48 psi.

With respect to the head-pressure relationship,
the following general statements can be made:

1. For any liquid of given and uniform
density, the pressure exerted by the liquid is
directly proportional to the head of the liquid.

2. At any given head, the pressure exerted by
any liquid is directly proportional to the density
of the liquid. Liquids having different densities
will exert different pressures at the same head.

15-3. Static and Velocity Heads. The total
head of any fluid is the sum of the static and
velocity heads of the fluid, viz:

h t =h s + h v (15-6)

where, h t = the total head in feet
h, = the static head in feet
h v = the velocity head in feet
The static head of any liquid is expressed as
the height in feet (or inches) of a gravitational
column of that liquid which would be required

276 PRINCIPLES OF REFRIGERATION

Static

n y pr T ure f\ |-t

Fig. 15-2. Illustrating relationship between the
static, velocity, and total pressures of a fluid flowing
in a circuit.

to produce a base pressure equal to the static
pressure of the liquid. That is, the head in feet
of liquid column equivalent to the static pressure
of the liquid is called the static head of the
liquid. Likewise, the head in feet of liquid
column equivalent to the velocity pressure of a
liquid is called the velocity head of the liquid.

The fundamental relationship between velo-
city and velocity head is established by Galileo's
law, which states in effect that all falling bodies,
regardless of weight, accelerate at equal rates
and that the final velocity of any falling body,
neglecting friction, depends only upon the height
from which the body falls. Hence, the height in
feet from which a body must fall in order to
attain a given velocity is the velocity head corre-
sponding to that velocity. The velocity head
corresponding to any given velocity can be
determined by applying the following equation:

h -*"

(15-7)

where, h v = the velocity head in feet

V = the velocity in feet per second (fps)
g = the acceleration due to gravity (32.2
ft/sec/sec)
By combining and/or rearranging Equations
15-7 and 15-4, the following relationships are
established:
To convert velocity head to velocity pressure,

JLxi
* - "14T (15 " 8)

To convert velocity to velocity pressure,
V* x P
Pv = Ig x 144

To convert velocity head to velocity,

V = y/2g x h v (15-10)

To convert velocity pressure to velocity,

fa * /». *

V P

144

(15-11)

(15-9)

15-4. Head-Energy Relationship. Although
the term "head" itself is entirely independent of
weight or density, it should be recognized that
the head of any fluid is numerically equal to the
energy per pound of fluid. For this reason, head
is often used to express energy per pound of
fluid.

The basic relationship of head to energy or
work is shown in the following equation :

Energy or work (ft-lb) = mass (lb) x head (ft)

(15-12)

Since velocity head (h v ) is equal to V*l7g
(Equation 15-7), it follows that the total velocity
(kinetic) energy (E„) of any given mass (Af ) of
fluid flowing at any given velocity (V) can be
expressed as

E k =Mx —

The fact that the preceding equation is iden-
tical to Equation 1-7 indicates that the velocity
head of a fluid is an expression of the kinetic
energy per pound of fluid. Similarly, it can be
shown also that the static head of a fluid is an
expression of the potential energy per pound of
fluid.

In any fluid column of uniform and constant
density, the potential energy per pound of fluid
is the same at all levels in the column. However,
the potential energy at various levels is differently
divided between the energy of position and the
energy of pressure (head) depending upon the
elevation. For example, in Fig. 15-1, 1 lb of
water at the uppermost level in the tank has a
potential energy of position with relation to the
base of 10 ft-lb (1 lb x 10 ft) in accordance with
Equation 1-8. Since the head at this level is
zero, the potential energy of pressure (head) is
also zero. On the other hand, 1 lb of water at
the base of the tank has no potential energy of
positions, but has pressure or head energy of
10 ft-lb (1 lb x 10 ft), according to Equation
15-12. Likewise, 1 lb of water at a level midway
in the water column also has potential energy in
the amount 10 ft-lb, the energy being evenly

FLUID FLOW, CENTRIFUGAL LIQUID PUMPS, WATER AND BRINE PIPING 277

divided between the energy of position and the
energy of pressure.

15-5. Static Head-Velocity Head Relation-
ship in Flowing Fluids. The fact, that the
static pressure of a fluid is exerted equally in all
directions, whereas the velocity pressure of the
fluid is exerted only in the direction of flow,
makes it relatively simple to measure the static
and velocity pressures (or heads) of a fluid
flowing in a conduit. This is illustrated in Fig.
15-2. Notice that tube A is so connected to the
conduit that the opening of the tube is exactly
perpendicular to the line of flow. Since only the
static pressure of the fluid will act in this direc-
tion, the height of the fluid column in tube A is
a measure of the static pressure or static head of
the fluid in the conduit. On the other hand,
tube B is so arranged in the conduit that the
opening of the tube is directly in the line of flow.
Since both the static pressure and the velocity
pressure of the flowing fluid act on the opening
of tube B, the height of the liquid column in tube
B is a measure of the total pressure or total head
of the fluid. Since the total pressure or head of a
fluid is the sum of the static and velocity
pressures or heads, it follows that the difference
in the heights of the two fluid columns is a
measure of the velocity pressure or velocity head
of the fluid in the conduit.

If losses because of friction are neglected, the
total pressure or head of a flowing fluid will be
the same at all points along the conduit. How-
ever, the total head may be differently divided
between static head and velocity head at the
several points, depending upon the velocity of
the fluid at these points.

For any given flow rate (quantity of flow), the
velocity of the fluid flowing in a conduit varies
inversely with the cross-sectional area of the
conduit. This relationship is expressed by the
basic equation

(15-13)

*-\

where V = the velocity in feet per second

Q = the flow rate in cubic feet per second
A = the cross-sectional area of the con-
duit in square feet

Note. When Q is in cubic feet per minute,
V will be in feet per minute.

In accordance with Equation 15-13, the fluid
velocity (and velocity head) in section B of the
conduit in Fig. 15-3 is greater than that in
sections A and C, since the cross-sectional area
of section B is less than that of sections A and C.
Assuming that the total head of the fluid is the
same at all points in the conduit, it follows then
that the static head-velocity head ratio in section
B is different from that in sections A and C. As
the fluid flows through the reducer between
sections A and B, static head is converted to
velocity head (pressure is converted to velocity).
Conversely, as the fluid flows through the
increaser between sections B and C, velocity
head is converted back into static head (velocity
is converted to pressure).

In view of the head-energy relationship, it is
evident that the conversion of static head to
velocity head is in fact a conversion of potential
energy (pressure) into kinetic energy (velocity).
Likewise, the conversion of velocity head to
static head represents a conversion of kinetic
energy (velocity) to potential energy (pressure).

Fig. 15-3. Illustrating changes in static-velocity pressure ratio resulting from changes in conduit area.

278 PRINCIPLES OF REFRIGERATION

15-6. Friction Head. It has already been
established that a fluid flowing in a conduit will
suffer losses in energy (converted into heat) as a
result of the work of overcoming friction. These
energy losses are frequently expressed in terms
of pressure drop or head loss. The pressure drop
in psi or the head loss in feet experienced by a
fluid flowing between any two points in a con-
duit is known as the friction head or friction loss
between these two points.

The amount of pressure drop or head loss
suffered by a fluid due to friction in flowing
through a conduit varies with a number of
factors: (1) the viscosity and specific gravity of
the fluid, (2) the velocity of the fluid, (3) the
hydraulic radius (ratio of perimeter to diameter)
of the conduit, (4) the roughness of the internal
surface of the conduit, and (5) the length of the
conduit.

Obviously, the mathematical evaluation of all
these factors is too laborious for most practical
purposes. As a general rule, the friction loss in
piping is determined from charts and tables.

The pressure (friction) loss in psi per hundred
feet of straight pipe is given in Charts 15-1 and
15-2 for various flow rates in various sizes of
pipe. Chart 15-1 applies to smooth copper tube,
whereas Chart 15-2 applies to fairly rough pipe.
Since the pressure loss for a given pipe size and
flow rate is proportional to the length of the
pipe, the pressure loss through any given length
of straight pipe is determined by the following
equation:

Total pressure loss (ft)

Total length of pipe (ft)
" 100

x pressure loss/100 ft (psi) (15-14)

Pipe fittings, such as elbows, tees, valves, etc.,
offer a greater resistance to flow than does
straight pipe and therefore must be taken into
account in determining the total friction loss
through the piping. For convenience, this is
done by considering the fittings as having a
resistance equal to a certain length of straight
pipe called the "equivalent length." Table 15-1
lists the equivalent length of straight pipe for
various types of fittings and valves. Notice that
the equivalent length varies with the size of the

Pipe-
When the equivalent length of the fittings is

added to the actual length of straight pipe, the

result is called the "total equivalent length."
This value is then applied in Equation 15-14 to
determine the total friction loss through the
piping.

Example 15-1. A water piping system con-
sists of 128 ft of 2 in. straight pipe, 6 standard
elbows, and 2 gate valves (full open). Using
fairly rough pipe, if the flow rate thrdugh the
system is 40 gpm, determine:

(a) The total equivalent length of straight pipe

(b) The total friction loss through the piping
in psi and in feet of water column.

Solution. From Table 15-1, the equivalent
lengths of 2 in. standard elbows and 2 in. gate
valves (full open) are 5 ft and 1 .2 ft, respectively.
From Chart 15-2, for a flow rate of 40 gpm, the
friction loss per hundred feet of 2 in. nominal
pipe is 3 psi. From Table 1-1, a pressure of
1 psi is equivalent to 2.31 ft of water column.
(a) Total equivalent length

Straight pipe = 128.0 ft

Six 2 in. elbows @ 5 ft = 30.0

Six 2 in. gate valves @
1.2 ft 2.4

160.4 ft

(b) Applying Equation
15-14, the total friction

loo XJ

loss through the piping

Converting to ft H 4

= 4.8 psi
= 4.8 x 2.31
= 11.09 ft H 2

Although the pressure loss determined from
Charts 15-1 and 15-2 apply only to water, the
charts can be used for other fluids by multiplying
the water pressure loss obtained from these
charts by the correction factors listed in Table
15-2.

15-7. Centrifugal Pumps. Liquid pumps used
in the refrigerating industry to circulate chilled
water or brine, and the condenser water are
usually of the centrifugal type.

A centrifugal pump consists mainly of a
rotating vane-type impeller that is enclosed
in a stationary casing. The liquid being pumped
is drawn in through the "eye" of the impeller
and is thrown to the outer edge or periphery of
the impeller by centrifugal force. Considerable
velocity and pressure are imparted to the liquid
in the process. The liquid leaving the periphery
of the impeller is collected in die casing and
directed through the discharge opening (Fig.
15-4).

FLUID FLOW, CENTRIFUGAL LIQUID PUMPS, WATER AND BRINE PIPING 279

Frequently, the impeller of the pump is
mounted directly on the shaft of the pump-
driving motor so that the pump and motor are
an integral unit (Fig. 1 5-5). In other cases, the
pump and motor are separate units and are
connected together by a. flexible coupling.

In general, the capacity of a centrifugal pump
depends on the design and size of the pump and
on the speed of the motor. For a pump of
specific size, design, and speed, the volume of
liquid handled varies with the pumping head

Flg> 15-4. Fluid flow through centrifugal pump.
(Courtejy Ingereoll-iUfld Company.)

against which the pump must wort. A charac-
teristic head-capacity curve for a typical centri-
fugal pump is shown in Fig. 1 5-6. Notice that
the pumping head is maximum when the valve
on the discharge of the pump is closed, at which
time the pump delivery is zero. As the valve is
opened, the pumping head decreases and the
deliver rate increases.

Centrifugal pumps are rated in gpm of
delivery at various pumping heads, that is,
centrifugal pumps are rated to deliver a certain
gpm against a certain pumping head. Although

Fig. 15-5, Typical centrifugal pump and motor
asig m bl y . (Gourteiy Bell & Getsttt Co m pany ,)

pump ratings are available in table form, more
frequently they are taken from head-capacity
curves (see Chart R-19). In either case, before
the proper pump can be selected from the manu-
facturer's ratings, it is necessary to know the
required gpm and the total pumping head
against which the pump must operate.
15-8. Total Pumping Head. The total pump-
ing head is the sum of the static head and the
friction head.

The static head is the vertical distance be-
tween the "free liquid level' 1 and the highest
point to which the liquid must be lifted by the
pump. For the condenser-water circulating sys-
tem in Fig. 15-7, the static bead, measured in
feet of water column. Is the vertical distance in
feet between the free water level in the tower
basin and the tower spray header. Because of
the water head in the tower basin, the water in
the discharge pipe will stand to the level of the

Gpm
Fig. 15-4. Centrifugal pump dalivery i
erewei h the pumping h«ad dttrn*«.
IngsrsoH-Rind Company.)

apacity in-
{CourtMy

280 PRINCIPLES OF REFRIGERATION

Total length of

straight pipe

80 ft

Gate valve >

Condenser

Static
head

Globe
/^valve

Pump-^
Fig. 15-7. Condenser-water circulating system.

and cooling towers, are found in the manu-
facturers' rating tables.

When more than one condenser (or chiller,
etc.) is used in the system, the condensers are
piped in parallel and only the condenser with the
largest pressure drop is considered in computing
the pumping head.

The pressure loss through the cooling tower,
as given by the tower manufacturer, is the total
head and includes both the tower static and
friction heads. Therefore, the static head of the
tower should not be considered separately in
determining the total pumping head. When the
tower static head is the only static head in the
system, the static head should be disregarded
entirely. However, in the event that an auxiliary
indoor storage tank is employed, as shown in
Fig. 14-24, the vertical distance between the
level of the water in the tank and the normal
water level in the tower basin must be treated
as a separate static head.

Since pump manufacturers always express the
pumping head in "feet of water column," it is
necessary to compute the pumping head in these
units. When the pressure loss through the
several system components is given in psi or in
other units of pressure, it must be converted to
feet of water column before it can be used in
computing the pumping head. The required
conversion factors are found in Table 1-1.

water in the tower basin of its own accord.
Therefore, the distance the water is actually
lifted by the pump is only the distance from the
water level in the tower basin up to the spray
header. Contrast this with the pumping system
shown in Fig. 15-8.

When the piping system is a closed circuit, as
in Fig. 1 5-9, there is no static head on the pump,
since the fluid on one side of the piping system
will exactly balance the fluid on the other side.

A typical piping system curve in which gpm
is plotted against total head is shown in Fig.
15-10. Notice that the total head increases as
the flow rate through the system increases and
that the increase in the total head results
entirely from an increase in the friction head,
the static head being constant.
15-9. Determining the Total Pumping
Head. The pressure loss through the various
system components, such as condensers, chillers,

Static
head

■* 1

Lower
tank

* | [|Purnp|

Fig. 15-8

Pipe system
curve

Friction loss

FLUID FLOW, CENTRIFUGAL LIQUID PUMPS, WATER AND BRINE PIPING 281

Example 15-2. The recirculating water sys-
tem shown in Fig. 15-7 is for a 10-ton refrig-
erating system. The flow rate over the tower is
40 gpm (4 gpm/ton). The flow rate through the
condenser is 30 gpm (3 gpm/ton), with 10 gpm
(1 gpm/ton) flowing through the condenser by-
pass. From the manufacturers' rating tables,
the tower head based on 4 gpm/ton is 24 ft of
water column, whereas toe pressure drop
through the condenser for 30 gpm is 11.2 psi or
25.9 ft of water column (11.2 x 2.31). If the
size of the piping is 2 in. nominal, determine the
total pumping head and select the proper pump
from Chart R-19.

Static head
(or elevation)

Solution. Total equiva-
lent length of pipe:
Straight pipe
3 — 2 in. standard el-
bows at 5 ft
2 — 2 in. tees (side out-
let) at 12 ft

4—2 in. gate valves
(open) at 1.2 ft

From Chart 15-2, the
pressure loss per 100 ft of
pipe (40 gpm and 2 in. pipe)

Applying Equation 15-14,
the total pressure loss
through the piping

Open balance
tank ""

80.0 ft

15.0

24.0

4.8
123.8 ft

3 psi
123.8

x 3

100
3.71 psi

Chilled water

air-cooling

coil

Water chiller

Fig. 15-9. Closed chilled water (or brine) circulating
system. To compute pumping head use circuit
having greatest friction loss. There is no static head.

Gpm

Fig. 15-10. Friction head of piping system increases
as flow rate through system increases. (Courtesy
Ingersoll-Rand Company.)

Converting to ft H a O

Total pumping head
Piping
Condenser
Tower

= 3.71 x 2.31
= 8.58 ft H 2

= 8.58 ft
= 25.90
= 24.00
= 57.58 ft H 2
From Chart R-19, select pump Model

#1531-28, which has a delivery capacity of 40

gpm at a 57-ft head.

Example 15-3. At the required flow rate of
100 gpm, a certain water system has a pumping
head of 60 ft of water column. Select the proper
pump from Chart R-19.

Solution. Reference to Chart R-19 shows that
pump Model #1531-30 is the smallest pump
which can be used. However, since this pump
will deliver 125 gpm at a 60-ft head, to obtain
the desired flow rate of 100 gpm, the pumping
head must be increased to 73 ft of water. This is
accomplished by throttling the pump with a
globe valve installed on the discharge side of the
pump. (The pump should never be throttled on
the suction side.)

15-10. Power Requirements. The power
required to drive the pump depends upon the
delivery rate in pounds per minute, the total
pumping head, and the efficiency of the pump,
viz:

Pounds per minute x total head in feet
" 33,000 x pump efficiency

Since the flow rate is usually in gpm, a more
practical equation is

Gpm x total head x 8.33 lb/gal

Bhp =■

33,000 x efficiency

282 PRINCIPLES OF REFRIGERATION

Combining constants,

Gpm x total head in feet
Bhp " ■ 3960 x efficiency (15 - 15)

Equation IS- IS applies to water. When a
liquid other than water is handled, the specific
gravity of the liquid must be taken into account,
viz:

Gpm x total head x specific gravity

Bhp =

3960 x efficiency

(15-16)

From Equation 15-16, it is evident that the
power required by the pump increases as the
delivery rate, total head, or specific gravity
increases, and decreases as the pump increases.

Typical pump horsepower and efficiency
curves are shown in Fig. 15-11. Notice that
pump horsepower is lowest at no delivery and
increases progressively as the delivery rate
increases. Hence, any decrease in the pumping
head will cause an increase in both the delivery
rate and the power requirements of the pump.

Pump efficiency, also lowest at no delivery,
increases to a maximum as the flow rate is
increased and then decreases as the flow rate is
further increased. The pump efficiency curve in
Fig. 15-11 indicates that the highest efficiency is
obtained when the pump is selected to deliver
the desired gpm when operating at some point
near the midpoint of its head-capacity curve.
15-11. Water Piping Design. In general, the
water piping should be designed for the mini-
mum friction loss consistent with reasonable
initial costs so that the pumping requirements
are maintained at a practical minimum. Water

Gpm

Fig. 15-1 1. Variations in pump horsepower and
efficiency with delivery rate. (Courtesy Ingersoll-
Rand Company.)

lines should be kept as short as possible and a
minimum amount of fittings should be used.

Standard weight steel pipe or Type "L"
copper tubing are usually employed for con-
denser water piping. Pipe sizes which will
provide water velocities in the neighborhood of
5 to 8 fps at the required flow rate will usually
prove to be the most economical. For example,
assume that 150 gpm of water are to be circu-
lated through 100 equivalent feet of piping. The
following approximate values of velocity and
friction loss are shown in Chart 15-2 for a flow
rate of 150 gpm through various sizes of pipe:

Pipe Size Velocity Friction Loss per 100 ft
(inches) (fps) (psi) (ftH a O)

2 15.5 31.5 72.8
2\ 10.0 10.5 24.3

3 7.1 4.8 11.1
3| 5.2 2.0 4.6

4 3.9 1.1 2.5

Notice that whereas increasing the pipe size
from 2 to 3 in. results in a considerable reduction
in the friction loss (from 72.8 to 11.1ft), a
further increase in the pipe size from 3 to 4 in.
reduces the friction loss by only an additional
9.4 ft of water column (11.1 to 2.54 ft), which
will not ordinarily justify the increase in the cost
of the pipe. Depending upon the characteristics
of the available pump, either 3 in. or 3 J in. pipe
should be used. For instance, assume two
separate systems having pumping heads, ex-
clusive of tiie friction loss in the piping, of 55 ft
of water column and 65 ft of water column,
respectively. Reference to Chart R-19 indicates
that the only suitable pump for either of the
systems is Model 31531-32, which has a delivery
rate of 150 gpm at a 70-ft head. Therefore, for
the system having the 55-ft head, the permissible
friction loss in the piping is 15 ft (70 — 55),
whereas for the system having the 65-ft head,
the permissible friction loss in the piping is only
5 ft (70 — 65). For the latter system, 3J in. pipe
must be used, since the use of 3 in. pipe would
result in a total pumping head in excess of the
allowable 70 ft and necessitate the use of the
next larger size pump. On the other hand, for
the former system, 3 in. pipe is the most prac-
tical size. The use of 3 J in. pipe in this instance
would result in a total pumping head of only
61 ft and would necessitate throttling of the

FLUID FLOW. CENTRIFUGAL LIQUID PUMPS, WATER AND BRINE PIPING 283

pump discharge in order to raise the pumping
head to 70 ft and obtain the desired flow rate of
ISO gpm.

In designing the piping system, care should be
taken to include all valves and fittings necessary
for the proper operation and maintenance of the
water circulating system. It is good practice to
install a globe valve on the discharge side of the
pump to regulate the water flow rate when the
latter is critical. Too, where the piping is long
and/or the quantity of water in the system is
large, shut-off valves installed on the inlet and
outlet of both the pump and the condenser will
permit repairs to these pieces of equipment
without the necessity of draining the tower. A
drain connection should be installed at the
lowest point in the piping and the piping should
be pitched downward so as to assure complete
drainage during winter shut down.

The pump must always be located at some
point below the level of the water in the tower
basin in order to assure positive and continuous
priming of the pump. When quiet operation is
required, the pump may be isolated from the
piping with short lengths of rubber hose. Auto-
mobile radiator hose is suitable for this purpose.

PROBLEMS

1. A water piping system consists of 13S ft of
2.5 in. nominal Type L smooth copper tube, 5
standard elbows, and 2 globe valves (full open).

If the flow rate through the pipe is 60 gpm,
determine:

(a) The total equivalent length of straight
pipe. Ans. 297.5 equivalent ft

(b) The total friction loss through the pipe is
feet of water column. Ans. 9.28 ft H 2

2. Rework Problem 1 using fairly rough pipe.

Ans. 12.02 ft H 2

3. A recirculating condenser water system con-
sists of 100 ft of straight pipe, and 6 standard
90° elbows. At the desired flow rate the pressure
drop through the condenser is 7.5 psi and the
pressure drop over the tower is 10 ft of water
column. If 60 gpm are circulated through the
system, determine:

(a) The total equivalent length of pipe.

Ans. 1 30 equivalent ft

(b) The total head against which the pump
must operate. Ans. 42.92 ft H 2

4. From the manufacturer's rating curves,
select a pump to fit the conditions of Problem 3.

5. For a Refrigerant- 12 system, select a water
regulating valve to meet the following con-
ditions:

(a) Desired condensing temperature range —

90° to 105°.
(6) Maximum entering water temperature —

85° F.

(d) Pressure available at city main during
period of peak loading — 50 psi.

(e) Pressure loss through condenser and water
piping — 12 psi.

Refrigerants

16-1. The Ideal Refrigerant. Generally speak-
ing, a refrigerant is any body or substance which
acts as a cooling agent by absorbing heat from
another body or substance. With regard to the
vapor-compression cycle, the refrigerant is the
working fluid of the cycle which alternately
vaporizes and condenses as it absorbs and gives
off heat, respectively. To be suitable for use as a
refrigerant in the vapor-compression cycle, a
fluid should possess certain chemical, physical,
and thermodynamic properties that make it both
safe and economical to use.

It should be recognized at the onset that there
is no "ideal" refrigerant and that, because of the
wide differences in the conditions and require-
ments of the various applications, there is no one
refrigerant that is universally suitable for all
applications. Hence, a refrigerant approaches
the "ideal" only to the extent that its properties
meet the conditions and requirements of the
application for which it is to be used.

Table 16-1 lists a number of fluids having
properties which render them suitable for use as
refrigerants.* However, it will be shown
presently that only a few of the more desirable
ones are actually employed as such. Some, used

* Since some of the fluorocarbon refrigerants,
first introduced to the industry under the trade
name "Freon," are now produced under several
different trade marks, the ASRE, in order to avoid
the confusion inherent in the use of either pro-
prietary or chemical names, has adopted a number-
ing system for the identification of the various
refrigerants. Table 16-1 lists the ASRE number
designation, along with the chemical name and
formula for each of the compounds listed.

extensively as refrigerants in the past, have been
discarded as more suitable fluids were developed.
Others, still in the development stage, show
promise for the future. Tables 16-2 through
16-6- list the thermodynamic properties of some
of the refrigerants in common use at the present
time. The use of these tables has already been,
described in an earlier chapter.
16-2. Safe Properties. Ordinarily, the safe
properties of the refrigerant are the prime con-
sideration in the selection of a refrigerant. It is
for this reason that some fluids, which otherwise
are highly desirable as refrigerants, find only
limited use as such. The more prominent of
these are ammonia and some of the straight
hydrocarbons.

To be suitable for use as a refrigerant, a fluid
should be chemically inert to the extent that it is
nonflammable, nonexplosive, and nontoxic both
in the pure state and when mixed in any propor-
tion with air. Too, the fluid should not react
unfavorably with the lubricating oil or with any
material normally used in the construction of
refrigerating equipment. Nor should it react
unfavorably with moisture which despite strin-
gent precautions is usually present at least to
some degree in all refrigerating systems. Fur-
thermore, it is desirable that the fluid be of such a
nature that it will not contaminate in any way
foodstuff or other stored products in the event
that a leak develops in the system.
16-3. Toxicity. Since all fluids other than air
are toxic in the sense that they will cause
suffocation when in concentrations large enough
to preclude sufficient oxygen to sustain life,
toxicity is a relative term which becomes mean-
ingful only when the degree of concentration
and the time of exposure required to produce
harmful effects are specified.

The toxicity of most commonly used refrig-
erants has been tested by National Fire Under-
writers. As a result, the various refrigerants are
separated into six groups according to their
degree of toxicity, the groups being arranged in
descending order (Column 1 of Table 16-7).
Those falling into Group 1 are highly toxic and
are capable of causing death or serious injury
in relatively small concentrations and/or short
exposure periods. On the other hand, those
classified in Group 6 are only mildly toxic, being
capable of causing harmful effects only in rela-
tively large concentrations. Since injury from

284

REFRIGERANTS 285

the latter group is caused more by oxygen
deficiency than by any harmful effects of the
fluids themselves, for all practical purposes the
fluids in Group 6 are considered to be nontoxic.
However, it should be pointed out that some
refrigerants, although nontoxic when mixed with
air in their normal state, are subject to decom-
position when they come in contact with an open
flame or an electrical heating element. The
products of decomposition thus formed are
highly toxic and capable of causing harmful
effects in small concentrations and on short
exposure. This is true of all the fluorocarbon
refrigerants (see Column 3 of Table 16-7).
16-4. Flammability and Explosiveness.
With regard to flammability and explosiveness,
most of the refrigerants in common use are
entirely nonflammable and nonexplosive
(Column 2 of Table 16-7). Notable exceptions
to this are ammonia and the straight hydro-
carbons. Ammonia is slightly flammable and
explosive when mixed in rather exact propor-
tions with air. However with reasonable
precautions, the hazard involved in using
ammonia as a refrigerant is negligible.

Straight hydrocarbons, on the other hand,
are highly flammable and explosive, and their
use as refrigerants except in special applications
and under the surveillance of experienced
operating personnel is not usually permissible.
Because of their excellent thermal properties,
the straight hydrocarbons are frequently
employed in ultra-low temperature applications.
In such installations, the hazard incurred by
their use is minimized by the fact that the
equipment is constantly attended by operating
personnel experienced in the use and handling
of flammable and explosive materials.

The "American Standard Safety Code for
Mechanical Refrigeration" sets forth in detail
the conditions and circumstances under which
the various refrigerants can be safely used.
Most local codes and ordinances governing
the use of refrigerating equipment are based on
this code, which is sponsored jointly by the
ASRE and ASA.

The degree of hazard incurred by the use of
toxic refrigerants depends upon a number of
factors, such as the quantity of refrigerant used
with relation to the size of the space into which
the refrigerant may leak, the type of occupancy,
whether or not open flames are present, the

odor of the refrigerant, and whether or not
experienced personnel are on duty to attend the
equipment. For example, a small quantity of
even a highly toxic refrigerant presents little
hazard when used in relatively large spaces in
that it is not possible in the event of a leak for
the concentration to reach a harmful level.
Too, the danger inherent in the use of toxic
refrigerants is somewhat tempered by the fact
that toxic refrigerants (including decomposition
products) have very noticeable odors which tend
to serve as a warning of their presence. Hence,
toxic refrigerants are usually a hazard only to
infants and others who, by reason of infirmity or
confinement, are unable to escape the fumes.
At the present time, ammonia is the only toxic
refrigerant that is used to any great extent, and
its use is ordinarily limited to packing plants,
ice plants, and large cold storage facilities where
experienced personnel are usually on duty.
16-5. Economic and Other Considerations.
Naturally, from the viewpoint of economical
operation, it is desirable that the refrigerant
have physical and thermal characteristics which
will result in the minimum power requirements
per unit of refrigerating capacity, that is, a high
coefficient of performance. For the most part,
the properties of the refrigerant which influence
the coefficient of performance are: (1) the
latent heat of vaporization, (2) the specific
volume of the vapor, (3) the compression ratio,
and (4) the specific heat of the refrigerant in
both the liquid and vapor states.

Except in very small systems, a high latent
heat value is desirable in that the weight of
refrigerant circulated per unit of capacity is
less. When a high latent heat value is accom-
panied by a low specific volume in the vapor
state, the efficiency and capacity of the com-
pressor are greatly increased. This tends not
only to decrease the power consumption but
also to reduce the compressor displacement
required, which permits the use of smaller, more
compact equipment. However, in small systems,
if the latent heat value of the refrigerant is too
high, the amount of refrigerant circulated will
be insufficient for accurate control of the liquid.

A low specific heat for the liquid and a high
specific heat for the vapor are desirable in that
both tend to increase the refrigerating effect
per pound, the former by increasing the sub-
cooling effect and the latter by decreasing the

286 PRINCIPLES OF REFRIGERATION

Refrigerent-50

Methane

(CH4) -259°F

Refrigerant-40
Methyl Chloride
(CH3CI) -11*F

Refrigerant-30
Methylene Chloride
(CH2CI2) 104"F

Refrigerant-20
Chloroform
(CHCI3) 142-F

Refrigerant-21
Dichloromonofluoromethane
(CHCI2F) 48'F

Refrigerant-22
Monochlorodifluoromethane
(CHCIF2) -41°F

Refrigerant-11
Trichloromonofluoromethane
(CClaF) 75'F

Refrigerant- 12
Dichlorodifluoromethane
(CCI2F2) -22°F

Refrigerant- 13
Monochlorotrifluoromethane
(CCIF3) — 115*F

Refrigerant-10
Carbontetrach loride
(CCI4) 169°F

Refrigerant- 14
Carbontetrafluoride
(CF 4 ) -198°F

Fig. 16-1. Methane series refrigerants.

REFRIGERANTS 287

superheating effect. When both are found in a
single fluid, the efficiency of a liquid-suction
heat exchanger is much improved.

The effect of compression ratio on the work
of compression and, consequently, on the
coefficient of performance, has already been
discussed in a previous chapter. Naturally, all
other factors being equal, the refrigerant giving
the lowest compression ratio is the most
desirable. Low compression ratios result in
low power consumption and high volumetric
efficiency, the latter being more important in
smaller systems since it permits the use of
small compressors.

Too, it is desirable that the pressure-tempera-
ture relationship of the refrigerant is such
that the pressure in the evaporator is always
above atmospheric. In the event of a leak on the
low pressure side of the system, if the pressure
in the low side is below atmospheric, consider-
able amounts of air and moisture may be drawn
into the system, whereas if the vaporizing
pressure is above atmospheric, the possibility of
drawing in air and moisture in the event of a
leak is minimized.

Reasonably low condensing pressures under
normal atmospheric conditions are also desirable
in that they allow the use of lightweight
materials in the construction of the condensing
equipment, thereby reducing the size, weight,
and cost of the equipment.

Naturally, the critical temperature and
pressure of the refrigerant must be above the
maximum temperature and pressure which will
be encountered in the system. Likewise, the
freezing point of the refrigerant must be safely
below the minimum temperature to be obtained
in the cycle. These factors are particularly
important in selecting a refrigerant for a low
temperature application.

In Table 16-8, a comparison is given of the
performance of the various refrigerants at
standard ton conditions (5° F evaporator and
86° F condensing). Notice particularly that,
with the exception of air, carbon dioxide, and
ethane, the horsepower required per ton of
refrigeration is very nearly the same for all the
refrigerants listed. For this reason, efficiency
and economy of operation are not usually
deciding factors in the selection of the refrig-
erant. More important are those properties
which tend to reduce the size, weight, and

initial cost of the refrigerating equipment and
which permit automatic operation and a
minimum of maintenance.
16-6. Early Refrigerants. In earlier days, when
mechanical refrigeration was limited to a
few large applications, ammonia and carbon
dioxide were practically the only refrigerants
available. Later, with the development of
small, automatic domestic and commercial
units, refrigerants such as sulfur dioxide and
methyl chloride came into use, along with
methylene chloride, which was developed for
use with centrifugal compressors. Methylene
chloride and carbon dioxide, because of their
safe properties, were extensively used in large
air conditioning applications.

With the exception of ammonia, all these
refrigerants have fallen into disuse and are
found only in some of the older installations,
having been discarded in favor of the more
suitable fluorocarbon refrigerants as the latter
were developed. The fluorocarbons are practi-
cally the only refrigerants in extensive use at the
present time. Again, an exception to this is
ammonia which, because of its excellent thermal
properties, is still widely used in such instal-
lations as ice plants, skating rinks, etc. A few
other refrigerants also find limited use in special
applications.

16-7. Development of the Fluorocarbons.
The search for a completely safe refrigerant
with good thermal properties led to the develop-
ment of the fluorocarbon refrigerants in the late
1920's. The fluorocarbons (fluoronated hydro-
carbons) are one group of a family of compounds
known as the halocarbons (halogenated hydro-
carbons). The halocarbon family of compounds
are synthesized by replacing one or more of the
hydrogen atoms in methane (CH*) or ethane
(CgHg) molecules, both of which are pure
hydrocarbons, with atoms of chlorine, fluorine,
and/or bromine, the latter group comprising the
halogen family. Halocarbons developed from
the methane molecule are known as "methane
series halocarbons." Likewise, those developed
from the ethane molecule are referred to as
"ethane series halocarbons."

The composition of the methane series halo-
carbons is shown in Fig. 16-1. Notice that the
basic methane molecule consists of one atom of
carbon (Q and four atoms of hydrogen (H).
If the hydrogen atoms are replaced progressively

288 PRINCIPLES OF REFRIGERATION

Refrigerant- 170

Ethane

(CH3CH3) -127.5-F

-113

Trichlorotrifluoroethane
(CCI2FCCIF2) 117.6'F

114

Dichlortetrafluofoethane
(CClFjjCClFii) 38.4°F

Fig. 16-2. Ethane series refrigerants.

with chlorine (CI) atoms, the resulting com-
pounds are methyl chloride (CH 3 C1), methylene
chloride (CHaClg), chloroform (CHCI3), and
carbontetrachloride (CC1 4 ), respectively, the
last two being the base molecules for the more
popular fiuorocarbons of the methane series.

If the chlorine atoms in the carbontetra-
chloride molecule are now replaced progressively
with fluorine atoms, the resulting compounds
are trichloromonofluoromethane (CC1 S F),
dichlorodifluoromethane (CCl a F 2 ), mono-
chlorotrifluoromethane (CCIF3), and carbon-
tetrafiuoride (CF^, respectively. In the same
order, the ASRE refrigerant standard number
designations for these compounds are Refrig-
erants-11, 12, 13, and 14, the last figure in the
numbers being an indication of the number of
fluorine atoms in the molecule.

The molecular structure of Refrigerants-21
and 22, which are also fiuorocarbons of the
methane series, is shown in Fig. 16-1. Notice
the presence of the hydrogen atom in each of
these two compounds, an indication that they
are derivatives of the chloroform molecule
rather than the carbontetrachloride molecule.

Figure 16-2 shows the molecular structure of
Refrigerants-113 and 114, the only two fiuoro-

carbons of the ethane series in common use.
The presence of the two carbon atoms identifies
the basic molecule as ethane, rather than
methane, which has only one carbon atom.
The individual characteristics of these and
other refrigerants are discussed in the following
sections.

16-8. The Effect of Moisture. It is a well-
established fact that moisture will combine in
varying degrees with most of the commonly
used refrigerants, causing the formation of
highly corrosive compounds (usually acids)
which will react with the lubricating oil and
with other materials in the system, including
metals. This chemical action often results in
pitting and other damage to valves, seals,
bearing journals, cylinder walls, and other
polished surfaces. It may also cause deteriora-
tion of the lubricating oil and the formation of
metallic and other sludges which tend to clog
valves and oil passages, score bearing surfaces,
and otherwise reduce the life of the equipment.
Moisture corrosion also contributes to com-
pressor valve failure and, in hermetic motor-
compressors, often causes breakdown of the
motor winding insulation, which results in
shorting or grounding of the motor.

Although a completely moisture-free refrig-
erating system is not possible, good refrig-
erating practice demands that the moisture
content of the system be maintained below the
level which will produce harmful effects in the
system. The minimum moisture level which
will produce harmful effects in a refrigerating
system is not clearly defined and will vary
considerably, depending upon the nature of the
refrigerant, the quality of the lubricating oil, and
the operating temperatures of the system, par-
ticularly the compressor discharge temperature.

Moisture in a refrigerating system may exist
as "free water" or it may be in solution with the
refrigerant. When moisture is present in the
system in the form of free water, it will freeze
into ice in the refrigerant control and/or in the
evaporator, provided that the temperature of
the evaporator is maintained below the freezing
point of the water. Naturally, the formation of
ice in the refrigerant control orifice will prevent
the flow of liquid refrigerant through that part
and render the system inoperative until such
time that the ice melts and flow through the
control is restored. In such cases, refrigeration
is usually intermittent as the flow of liquid is
started and stopped by alternate melting and
freezing of the ice in the control orifice.

Since free water exists in the system only
when the amount of moisture in the system
exceeds the amount that the refrigerant can
hold in solution, freeze-ups are nearly always
an indication that the moisture content of the
system is above the minimum level that will
produce corrosion. On the other hand, the
mere absence of freeze-ups cannot be taken to
mean that the moisture content of the system
is necessarily below the level which will cause
corrosion, since corrosion can occur with some
refrigerants at levels well below those which
will result in free water. Too, it must be
recognized that freeze-ups do not occur in air
conditioning systems or in any other system
where the evaporator temperature is above
the freezing point of water. For this reason,
high temperature systems are often more sub-
ject to moisture corrosion than are systems
operating at lower evaporator temperatures,
since relatively large quantities of moisture can
go unnoticed in such systems for relatively long
periods of time.

Since the ability of an individual refrigerant

REFRIGERANTS 289

to hold moisture in solution decreases as the
temperature decreases, it follows that the
moisture content in low temperature systems
must be maintained at a very low level in order to
avoid freeze-ups. Hence, moisture corrosion in
low temperature systems is usually at a minimum.

The various refrigerants differ greatly both as
to the amount of moisture they will hold in
solution and as to the effect that the moisture
has upon them. For example, the straight
hydrocarbons, such as propane, butane, ethane,
etc., absorb little if any moisture. Therefore,
any moisture contained in such systems will be
in the form of free water and will make its
presence known by freezing out in the refrig-
erant control. Since this moisture must be
removed immediately in order to keep the system
operative, moisture corrosion will not usually
be a problem when these refrigerants are used.

Ammonia and sulfur dioxide, on the other
hand, have an affinity for water and therefore
are capable of absorbing moisture in such large
quantities that free water is seldom found
in systems employing these two refrigerants.
However, the effects produced by the combina-
tion of the water and the refrigerant are entirely
different for the two refrigerants.

In ammonia systems, the combination of
water and ammonia produces aqua ammonia, a
strong alkali, which attracts nonferrous metals,
such as copper and brass, but has little if any
effect on iron or steel or any other materials in
the system. For this reason, ammonia systems
can be operated successfully even when rela-
tively large amounts of moisture are present in
the system.

In the case of sulfur dioxide, the moisture
and sulfur dioxide combine to form sulfurous
acid (H2SO3), which is highly corrosive. In
view of the high solubility of water in S0 2 , the
amount of acid formed can be quite large.
Hence, corrosion in sulfur dioxide systems can
be very heavy.

The halocarbon refrigerants hydrolyze only
slightly and therefore form only small amounts
of acids or other corrosive compounds. As a
general rule, corrosion will not occur in systems
employing halocarbon refrigerants when the
moisture content is maintained below the level
which will cause freeze-ups, provided that high
quality lubricating oils are used and that dis-
charge temperatures are reasonably low.

290 PRINCIPLES OF REFRIGERATION

16-9. Refrigerant-Oil Relationship. With a
few exceptions, the oil required for lubrication
of the compressor is contained in the crankcase
of the compressor where it is subject to contact
with the refrigerant. Hence, as already stated,
the refrigerant must be chemically and physically
stable in the presence of oil, so that neither the
refrigerant nor the oil is adversely affected by
the relationship.

Although some refrigerants, particularly sul-
fur dioxide and the halocarbons, react with
the lubricating oil to some extent, under normal
operating conditions the reaction is usually
slight and therefore of little consequence,
provided that a high quality lubricating oil is
used and that the system is relatively clean and
dry. However, when contaminants, such as air
and moisture, are present in the system in any
appreciable amount, chemical reactions involv-
ing the contaminants, the refrigerant, and the
lubricating oil often occur which can result in
decomposition of the oil, the formation of
corrosive acids and sludges, copper plating,
and/or serious corrosion of polished metal
surfaces. High discharge temperatures greatly
accelerate these processes, particularly oil de-
composition, and often result in the formation of
carbonaceous deposits on discharge valves and
pistons and in the compressor head and dis-
charge line. This condition is aggravated by the
use of poorly refined lubricating oils containing
a high percentage of unsaturated hydrocarbons,
the latter being very unstable chemically.

Because of the naturally high discharge
temperature of Refrigerant-22 (see Table 16-8),
breakdown of the lubricating oil, accompanied
by motor burnouts, is a common problem with
hermetic motor-compressor units employing
this refrigerant, particularly when used in con-
junction with air-cooled condensers and long
suction lines.

Copper plating of various compressor parts
is often found in systems employing halocarbon
refrigerants. The parts usually affected are the
highly polished metal surfaces which generate
heat, such as seals, pistons, cylinder walls,
bearing surfaces, and valves. The exact cause of
copper plating has not been definitely deter-
mined, but considerable evidence does exist that
moisture and poor quality lubricating oils
are contributing factors.

Because copper is never used with ammonia,

copper plating is not found in ammonia systems.
However, neither is it found in sulfur dioxide
systems, although copper has been employed
extensively with this refrigerant.

In any event, regardless of the nature of
and/or the cause of unfavorable reactions
between the refrigerant and the lubricating oil,
these disadvantages can be greatly minimized
or eliminated by the use of high quality lubri-
cating oils, having low "pour" and/or "floe"
points (see Section 18-16), by maintaining the
system relatively free of contaminants, such as
air and moisture, and by designing the system so
that discharge temperatures are reasonably low.
16-10. Oil Miscibility. With regard to the
refrigerant-oil relationship, one important
characteristic which differs for the various
refrigerants is oil miscibility, that is, the ability
of the refrigerant to be dissolved into the oil and
vice versa.

With reference to oil miscibility, refrigerants
may be divided into three groups: (1) those
which are miscible with oil in all proportions
under conditions found in the refrigerating
system, (2) those which are miscible under
conditions normally found in the condensing
section, but separate from the oil under the
conditions normally found in the evaporator
section, and (3) those which are not miscible
with oil at all (or only very slightly so) under
conditions found in the system.

As to whether or not oil miscibility is a
desirable property in a refrigerant there is some
disagreement. In any event, the fact of oil
miscibility, or the lack of it, has little if any
significance insofar as the selection of the
refrigerant is concerned. However, since it
greatly influences the design of the compressor
and other system components, including the
refrigerant piping, the degree of oil miscibility
is an important refrigerant characteristic and
therefore should be considered in some detail.

With regard to the oil, one of the principal
effects of an oil miscible refrigerant is to dilute
the oil in the crankcase of the compressor,
thereby lowering the viscosity (thinning) of the
oil and reducing its lubricating qualities. To
compensate for refrigerant dilution, the com-
pressor lubricating oil used in conjunction with
oil-miscible refrigerants should have a higher
initial viscosity than that used for similar duty
with nonmiscible refrigerants.

REFRIGERANTS 291

Viscosity may be defined as a measure of
fluid friction or as a measure of the resistance
that a fluid offers to flow. Hence, thin, low
viscosity fluids will flow more readily than
thicker, more viscous fluids. To provide
adequate lubrication for the compressor, the
viscosity of the lubricating oil must be main-
tained within certain limits. If the viscosity of
the oil is too low, the oil will not have sufficient
body to form a protective film between the
various rubbing surfaces and keep them
separated. On the other hand, if the viscosity
of the oil is too high, the oil will not have
sufficient fluidity to penetrate between the
rubbing surfaces, particularly where tolerances
are close. In either case, lubrication of the
compressor will not be adequate.

Any oil circulating through the system with
the refrigerant will have an adverse affect on
the efficiency and capacity of the system, the
principal reason being that the oil tends to
adhere to and to form a film on the surface of
the condenser and evaporator tubes, thereby
lowering the heat transfer capacity of these two
units. Since the oil becomes more viscous and
tends to congeal as the temperature is reduced,
the problem with oil is greatest in the evaporator
and becomes more acute as the temperature of
the evaporator is lowered.

Since the only reason for the presence of oil
in the refrigerating system is to lubricate the
compressor, it is evident that the oil will best
serve its function when confined to the com-
pressor and not allowed to circulate with the
refrigerant through other parts of the system.
However, since, with few exceptions, the system
refrigerant unavoidably comes into contact with
the oil in the compressor, a certain amount of
oil in the form of small particles will be en-
trained in the refrigerant vapor and carried over
through the discharge valves into the discharge
line. If the oil is not removed from the vapor
at this point, it will pass into the condenser and
liquid receiver from where it will be carried
to the evaporator by the liquid refrigerant.
Obviously, in the interest of system efficiency
and in order to maintain the oil in the crank-
case at a constant level, some provision must be
made for removing this oil from the system and
returning it to the crankcase where it can per-
form its lubricating function.

The degree of difficulty experienced in bring-

ing about the return of oil to the crankcase
depends primarily on three factors: (1) the
oil miscibility of the refrigerant, (2) the type of
evaporator used, and (3) the evaporator tem-
perature.

When an oil-miscible refrigerant is employed,
the problem of oil return is greatly simplified
by the fact that the oil remains in solution with
the refrigerant. This permits the oil to be
carried along through the system by the
refrigerant and, subsequently, to be returned to
the crankcase through the suction line, provided
that the evaporator and the refrigerant piping
are properly designed.

Unfortunately, when nonmiscible refriger-
ants are used, once the oil passes into the
condenser, the return of the oil to the crankcase
is not so easily accomplished. The reason for
this is that, except for a small amount of
mechanical mixing, the refrigerant and the oil
will remain separate, so that only a small
portion of the oil is actually carried along with
the refrigerant. For example, in the case of
ammonia, which is lighter than oil, a large
percentage of the oil will separate from the
liquid ammonia and settle out at various low
points in the system. For this reason, oil
drains should be provided at the bottom of all
receivers, evaporators, accumulators, and other
vessels containing liquid ammonia, and pro-
visions should be made for draining the oil
from these points, either continuously or
periodically, and returning it to the crankcase.
This may be accomplished manually or auto-
matically.

When flooded-type evaporators are used, the
refrigerant velocity will not usually be sufficient
to permit the refrigerant vapor to entrain the
oil and carry it over into the suction line and
back to the crankcase. Hence, even with oil
miscible refrigerants, where flooded-type evap-
orators are employed, it is often necessary to
make special provisions for oil return. The
methods used to insure the continuous return
of the oil from the evaporator to the crankcase
in such cases is described in Chapter 19.

Since the oil acts to lubricate the refrigerant
flow control and other valves which may be in
the system, the circulation of a small amount of
oil with the refrigerant is not ordinarily objec-
tionable. However, because of the adverse
effect on system capacity, the amount of oil

292 PRINCIPLES OF REFRIGERATION

should be kept to a practical minimum. Too,
since the oil in circulation comes initially from
the compressor crankcase, an excessive amount
in circulation may cause the oil level in the
crankcase to fall below the minimum level
required for adequate lubrication of the com-
pressor parts.

In order to minimize the circulation of oil, an
oil separator or trap is sometimes installed
in the discharge line between the compressor
and the condenser (see Section 19-12).

As a general rule, discharge line oil separators
should be employed in any system where oil
return is likely to be inadequate and/or where
the amount of oil in circulation is apt to be
excessive or to cause an undue loss in system
capacity and efficiency. Specifically, discharge
line oil separators are recommended for all
systems employing nonmiscible refrigerants
(or refrigerants which are not oil miscible at the
evaporator conditions), not only because of the
difficulty experienced in returning the oil from
the evaporator to the crankcase but also because
the presence of even small amounts of oil in the
evaporators of such systems will usually cause
considerable loss of evaporator efficiency and
capacity.

The same thing is usually true for systems
employing miscible refrigerants when the
evaporator temperature is below 0° F. Oil
separators are recommended also for all
systems using flooded evaporators, since oil
return from this type of evaporator is apt to
be inadequate because of low refrigerant
velocities.

Although oil separators are very effective in
removing oil from the refrigerant vapor, they
are not 100% efficient. Therefore, even though
an oil separator is used, some means must still
be provided for returning to the crankcase the
small amount of oil which will always pass
through the separator and find its way into
other parts of the system. Too, since oil
separators can often cause serious problems in
the system if they are not properly installed,
the use of oil separators should ordinarily be
limited to those systems where the nature of the
refrigerant or the particular design of the
system requires their use. Oil separators are
discussed in more detail in Chapter 19.
16-11. Leak Detection. Leaks in a refrigerat-
ing system may be either inward or outward,

depending on whether the pressure in the
system at the point of leakage is above or below
atmospheric pressure. When the pressure in
the system is above atmospheric at the point of
leakage, the refrigerant will leak from the
system to the outside. On the other hand,
when the pressure in the system is below
atmospheric, there is no leakage of refrigerant
to the outside, but air and moisture will be
drawn into the system. In either case, the
system will usually become inoperative in a
very short time. However, as a general rule,
outward leaks are less serious than inward ones,
usually requiring only that the leak be found and
repaired and that the system be recharged with
the proper amount of refrigerant. In the case of
inward leaks, the air and moisture drawn into
the system increase the discharge pressure and
temperature and accelerates the rate of corro-
sion. The presence of moisture in the system
may also cause freeze-up of the refrigerant
control. Furthermore, after the leak has been
located and repaired, the system must be
completely evacuated and dehydrated before it
can be placed in operation. A refrigerant drier
should also be installed in the system.

The necessity of maintaining the system free
of leaks demands some convenient means for
checking a new system for leaks and for
detecting leaks if and when they occur in
systems already in operation. New systems
should be checked for leaks under both vacuum
and pressure.

One method of leak detection universally
used with all refrigerants employs a relatively
viscous soap solution which is relatively free of
bubbles. The soap solution is first applied to
the pipe joint or other suspected area and then
examined with the help of a strong light. The
formation of bubbles in the soap solution
indicates the presence of a leak. For adequate
testing with a soap solution, the pressure in the
system should be 50 psig or higher.

The fact that sulfur and ammonia vapors
produce a dense white smoke (ammonia
sulfite) when they come into contact with one
another provides a convenient means of
checking for leaks in both sulfur dioxide and
ammonia systems. To check for leaks in a
sulfur dioxide system, a cloth swab saturated
with stronger ammonia (approximately 28%
available in any drug store) is held near, but

not in contact with, all pipe joints and other
suspected areas. A leak is indicated when the
ammonia swab gives off a white smoke.

Ammonia systems are checked in the same
way except that a sulfur candle is substituted for
the ammonia swab. Dampened phenophthalein
paper, which turns red on contact with am-
monia vapor, may also be used to detect
ammonia leaks.

A halide torch is often used to detect leaks in
systems employing any of the halocarbon refrig-
erants. The halide torch consists of a copper
element which is heated by a flame. Air to
support combustion is drawn in through a
rubber tube, one end of which is attached to the
torch. The free end of the tube is passed around
all suspected areas. The presence of a halo-
carbon vapor is indicated when the flame
changes from its normal color to a bright green
or purple. The halide torch should be used only
in well-ventilated spaces.

For carbon dioxide and the straight hydro-
carbons, the only method of leak detection is
the soap solution previously mentioned.
16-12. Ammonia. Ammonia is the only refrig-
erant outside of the fluorocarbon group that is
being used to any great extent at the present
time. Although ammonia is toxic and also
somewhat flammable and explosive under cer-
tain conditions, its excellent thermal properties
make it an ideal refrigerant for ice plants,
packing plants, skating rinks, large cold storage
facilities, etc., where experienced operating
personnel are usually on duty and where its
toxic nature is of little consequence.

Ammonia has the highest refrigerating effect
per pound of any refrigerant. This, together
with a moderately low specific volume in the
vapor state, makes possible a high refrigerating
capacity with a relatively small piston dis-
placement.

The boiling point of ammonia at standard
atmospheric pressure is -28° F. The evapora-
tor and condenser pressures at standard ton
conditions of 5° F and 86° F are 19.6 psig and
154.5 psig, respectively, which are moderate, so
that lightweight materials can be used in the
construction of the refrigerating equipment.
However, the adiabatic discharge temperature
is relatively high, being 210° F at standard ton
conditions, which makes water cooling of the
compressor head and cylinders desirable. Too,

REFRIGERANTS 293

high suction superheats should be avoided in
ammonia systems.

Although pure anhydrous ammonia is non-
corrosive to all metals normally used in refrig-
erating systems, in the presence of moisture,
ammonia becomes corrosive to nonferrous
metals, such as copper and brass. Obviously,
these metals should never be used in ammonia
systems.

Ammonia is not oil miscible and therefore
will not dilute the oil in the compressor crank-
case. However, provisions must be made for
the removal of oil from the evaporator and an
oil separator should be used in the discharge line
of all ammonia systems.

Ammonia systems may be tested for leaks
with sulfur candles, which give off a dense white
smoke in the presence of ammonia vapor, or by
applying a thick soap solution around the pipe
joints, in which case a leak is indicated by the
appearance of bubbles in the solution.
16-13. Sulfur Dioxide. Sulfur dioxide (SO s )
is produced from the combustion of sulfur. It
is highly toxic, but nonflammable and
nonexplosive. In the 1920s and 1930s, sulfur
dioxide was widely used in domestic refrigerators
and in small commercial fixtures. Today, it is
found only in a few of the older commercial
units, having been replaced first by methyl
chloride and later by the more desirable fluoro-
carbon refrigerants.

The boiling point of sulfur dioxide at atmos-
pheric pressure is approximately 14° F. Satur-
ation pressures at standard ton conditions of
5°F and 86° F are 5.9 in. Hg and 51.8 psig,
respectively.

Sulfur dioxide is not oil miscible. However,
unlike ammonia and carbon dioxide, liquid sul-
fur dioxide is heavier than oil so that the oil
floats on top of the refrigerant. Since this
characteristic simplifies the problem of oil
return, it accounts for the popularity enjoyed by
sulfur dioxide in the past for small automatic
equipment.

Like most common refrigerants, sulfur dioxide
in the pure state is noncorrosive to metals
normally used in the refrigerating system. How-
ever, it combines with moisture to form
sulfurous acid (H 2 SOg) and sulfuric acid
(HgSO^, both of which are highly corrosive.
16-14. Carbon Dioxide. Carbon dioxide (COj)
is one of the first refrigerants used in mechanical

294 PRINCIPLES OF REFRIGERATION

refrigerating systems. It is odorless, nontoxic,
nonflammable, nonexplosive, and noncorrosive.
Because of its safe properties, it has been widely
used in the past for marine service and for air
conditioning in hospitals, theaters, hotels, and
in other places where safety is the prime con-
sideration. Although a few of these older
installations are still in service, at the present
time the use of carbon dioxide as a refrigerant
is limited for the most part to extremely low
temperature applications, particularly in the
production of solid C0 2 (dry ice).

One of the chief disadvantages of carbon
dioxide is its high operating pressures, which
under standard ton conditions of 5° F and 86° F
are 317.5 psig and 1031 psig, respectively.
Naturally, this requires the use of extra heavy
piping and equipment. However, because of the
high vapor density of CO a , the volume of vapor
handled by the compressor is only 0.96 cu ft per
minute per ton at 5° F, so that compressor sizes
are small.

Another disadvantage of carbon dioxide is
that the horsepower required per ton is approxi-
mately twice that of any of the commonly used
refrigerants. For carbon dioxide, the theoretical
horsepower required per ton at standard con-
ditions is 1.84, whereas for ammonia, the horse-
power required per ton is only 0.989, the latter
value being typical for most refrigerants.

Since its boiling temperature at atmospheric
pressure ( — 109.3° F) is below its freezing tem-
perature (— 69.9° F) at this pressure, carbon
dioxide cannot exist in the liquid state at atmos-
pheric pressure nor at any pressure below its
triple point pressure of 75.1 psia. At any
pressure under 75.1 psia, solid carbon dioxide
sublimes directly into the vapor state and there-
fore below this pressure is found only in the
solid and vapor states. Because of the low
critical temperature of CO a (87.8° F), relatively
low condensing temperatures are required for
liquefaction. Carbon dioxide is nonmiscible in
oil and therefore will not dilute the oil in the
crankcase of the compressor. Like ammonia,
it is lighter than oil. Hence, oil return problems
are similar to those encountered in an ammonia
system.

Leak detection is by soap solution only.
16-15. Methyl Chloride. Methyl chloride
(CH 3 C1) is a halocarbon of the methane series.
It has many of the properties desirable in a

refrigerant, which accounts for its wide use in
the past in both domestic and commercial appli-
cations. Its boiling point at atmospheric
pressure is —10.65° F. Evaporator and con-
denser pressures at standard ton conditions are
6.5 psig and 80 psig, respectively.

Although methyl chloride is considered non-
toxic, in large concentrations it has an anesthetic
effect similar to that of chloroform, a compound
to which it is closely related. Methyl chloride is
moderately flammable and is explosive when
mixed with air in concentrations between 8.1
and 17.2% by volume. The hazard resulting
from these properties is the principal reason for
the discarding of methyl chloride in favor of the
safer fluorocarbon refrigerants.

Methyl chloride is corrosive to aluminum,
zinc, and magnesium, and the compounds
formed in combination with these materials are
both flammable and explosive. Hence, these
metals should not be used in methyl chloride
systems. In the presence of moisture, methyl
chloride forms a weak hydrochloric acid, which
is corrosive to both ferrous and nonferrous
metals. Too, since natural rubber and the
synthetic, Neoprene, are dissolved by methyl
chloride, neither is suitable gasket material for
use in methyl chloride systems.

Oil return in methyl chloride systems is
simplified by the fact that methyl chloride is oil
miscible. However, in selecting the compressor
lubricating oil, crankcase dilution must be taken
into account.

Leaks in a methyl chloride system are found
with the aid of a soap solution which is applied
to the suspected joints. The presence of methyl
chloride vapor may be detected with a halide
leak detector. However, this method is not
recommended because of the flammability of
methyl chloride.

16-16. Methylene Chloride (Carrene I).
Methylene chloride (CH 2 C1 2 ), another halo-
carbon of the methane series, has a boiling point
of 103.5° F at atmospheric pressure, a charac-
teristic which permits the refrigerant to be stored
in sealed cans rather than in compressed gas
cylinders. Under standard ton conditions, the
evaporator and condenser pressures are both
below atmospheric pressure, being 27.6 in. Hg
and 9.5 in. Hg, respectively. Since the volume
of the vapor handled per ton of refrigerating
capacity is quite large (74.3 cu ft/min/ton at

REFRIGERANTS 295

5° F), centrifugal compressors, which are par-
ticularly suited to handling large volumes of low
pressure vapor, are required.

Although it dissolves natural rubber, methy-
lene chloride is noncorrosive even in the presence
of moisture. It is also nontoxic and nonflam-
mable. Because of its safe properties, it has been
widely used in large air conditioning installations.

The fact that methylene chloride is oil miscible
is of little consequence, since in centrifugal com-
pressors the oil and refrigerant do not ordinarily
come in contact with one another.

A halide torch or soap solution may be used
to detect leaks. However, the pressure in the
system must be built up above atmospheric in
either case.

16-17. Refrigerant-ll. Refrigerant- 1 1 (CC1 3 F)
is a fluorocarbon of the methane series and
has a boiling point at atmospheric pressure of
74.7° F. Operating pressures at standard ton
conditions are 24 in. Hg and 3.6 psig, respec-
tively, which is very similar to those of methylene
chloride. Although the theoretical horsepower
required at standard ton conditions (0.927) is
approximately the same as that for methylene
chloride, the compressor displacement required
at these conditions (36.32 cu ft/min/ton) is only
approximately one-half that required for methy-
lene chloride.

Like other fluorocarbon refrigerants, Refrig-
erant- 11 dissolves natural rubber. However,
it is noncorrosive, nontoxic, and nonflammable.
The low operating pressures and the relatively
high compressor displacement required necessi-
tate the use of a centrifugal compressor.

Refrigerant-ll is used mainly in the air
conditioning of small office buildings, factories,
department stores, theaters, etc. A halide torch
may be used for leak detection.
16-18. Refrigerant- 1 2. Although its suprem-
acy is being seriously challenged in some areas
by Refrigerant-22, Refrigerant- 12 (CCl 8 Fg) is by
far the most widely used refrigerant at the
present time. It is a completely safe refrigerant
in that it is nontoxic, nonflammable, and non-
explosive. Furthermore, it is a highly stable
compound which is difficult to break down even
under extreme operating conditions. However,
if brought into contact with an open flame or
with an electrical heating element, Refrigerant-
12 will decompose into products which are
highly toxic (see Section 16-3).

Along with its safe properties, the fact that
Refrigerant- 12 condenses at moderate pressures
under normal atmospheric conditions and has a
boiling temperature of — 21 °F at atmospheric
pressure makes it a suitable refrigerant for use
in high, medium, and low temperature appli-
cations and with all three types of compressors.
When employed in conjunction with multistage
centrifugal type compressors, Refrigerant-12
has been used to cool brine to temperatures as
low as -110° F.

The fact that Refrigerant-12 is oil miscible
under all operating conditions not only sim-
plifies the problem of oil return but also tends
to increase the efficiency and capacity of the
system in that the solvant action of the refrig-
erant maintains the evaporator and condenser
tubes relatively free of oil films which otherwise
would tend to reduce the heat transfer capacity
of these two units.

Although the refrigerating effect per pound for
Refrigerant-12 is relatively small as compared to
that of some of the other popular refrigerants,
this is not necessarily a serious disadvantage.
In fact, in small systems, the greater weight of
Refrigerant-12 which must be circulated is a
decided advantage in that it permits closer
control of the liquid. In larger systems, the
disadvantage of the low latent heat value is
offset somewhat by a high vapor density, so that
the compressor displacement required per ton of
refrigeration is not much greater than that
required for Refrigerants-22, 500, and 717. The
horsepower required per ton of capacity com-
pares favorably with that required for other
commonly used refrigerants.

A halide torch is used for leak detection.
16-19. Refrigerant- 1 3. Refrigerant-13(CC1F,)
was developed for and is being used in ultra-low
temperature applications, usually in the low
stage of a two or three stage cascade system. It
is also being used to replace Refrigerant-22 in
some low temperature applications.

The boiling temperature of Refrigerant- 13 is
— 144.5° F at atmospheric pressure. Evaporator
temperatures down to — 150°F are practical.
The critical temperature is 83.9° F. Since con-
densing pressures and the compressor displace-
ment required are both moderate, Refrigerant-1 3
is suitable for use with all three types of
compressors.

Refrigerant-13 is a safe refrigerant. It is not

296 PRINCIPLES OF REFRIGERATION

miscible with oil. A halide torch may be used
for leak detection.

16-20. Refrigerant-22. Refrigerant-22
(CHClFj) has a boiling point at atmospheric
pressure of —41.4° F. Developed primarily as a
low temperature refrigerant, it is used exten-
sively in domestic and farm freezers and in
commercial and industrial low temperature
systems down to evaporator temperatures as
low as — 125°F. It also finds wide use in
packaged air conditioners, where, because of
space limitations, the relatively small compressor
displacement required is a decided advantage.

Both the operating pressures and the adiabatic
discharge temperature are higher for Refrig-
erant-22 than for Refrigerant-12. Horsepower
requirements are approximately the same.

Because of the high discharge temperatures
experienced with Refrigerant-22, suction super-
heat should be kept to a minimum, particularly
where hermetic motor-compressors are em-
ployed. In low temperature applications, where
compression ratios are likely to be high, water
cooling of the compressor head and cylinders is
recommended in order to avoid overheating of
the compressor. Air-cooled condensers used
with Refrigerant-22 should be generously sized.

Although miscible with oil at temperatures
found in the condensing section, Refrigerant-22
will often separate from the oil in the evaporator.
The exact temperature at which separation
occurs varies considerably with the type of oil
and the amount of oil mixed with the refrigerant.
However, no difficulty is usually experienced
with oil return from the evaporator when a
properly designed serpentine evaporator is used
and when the suction piping is properly designed.
When flooded evaporators are employed, oil
separators should be used and special pro-
visions should be made to insure the return of
oil from the evaporator. Oil separators should
always be used on low temperature applica-
tions.

The principal advantage of Refrigerant-22
over Refrigerant-12 is the smaller compressor
displacement required, being approximately
60% of that required for Refrigerant-12. Hence,
for a given compressor displacement, the refrig-
erating capacity is approximately 60% greater
with Refrigerant-22 man with Refrigerant-12.
Too, refrigerant pipe sizes are usually smaller
for Refrigerant-22 than for Refrigerant-12. For

evaporator temperatures between —20 and
—40° F, still another advantage added to
Refrigerant-22 is that the evaporator pressures
for Refrigerant-22 at these temperatures are
above atmospheric, whereas for Refrigerant-12
the evaporator pressures will be below atmos-
pheric. However, all this should not be taken to
mean that Refrigerant-22 is superior to Refrig-
erant-12 in all applications. As a matter of fact,
except in those applications where space limita-
tions necessitate the use of the smallest possible
equipment and/or where the evaporator tem-
perature is between -20° F and -40° F,
Refrigerant-12, because of its lower discharge
temperatures and greater miscibility with oil, is
probably the more desirable of the two
refrigerants.

The ability of Refrigerant-22 to absorb mois-
ture is considerably greater than that of Refrig-
erant-12 and therefore less trouble is experi-
enced with freeze-ups in Refrigerant-22 systems.
Although some consider this to be an advantage,
the advantage gained is questionable, since any
amount of moisture in a refrigerating system is
undesirable.

Being a fluorocarbon, Refrigerant-22 is a safe
refrigerant. A halide torch may be used for leak
detection.

16-21. Refrigerant-1 13. Refrigerant-113
(CCljjFCClFa) boils at 117.6° F under atmos-
pheric pressure. Operating pressures at standard
ton conditions are 27.9 in. Hg and 13.9 in. Hg,
respectively. Although the compressor displace-
ment per ton is somewhat high ( 1 00.76 cu ft/min/
ton at standard ton conditions), the horsepower
required per ton compares favorably with other
common refrigerants. The low operating
pressures and the large displacement required
necessitate the use of a centrifugal type
compressor.

Although used mainly in comfort air con-
ditioning applications, it is also employed in
industrial process water and brine chilling down
to 0° F.

Refrigerant-1 13 is a safe refrigerant. A halide
torch may be used for leak detection.
16-22. Refrigerant-1 14. Refrigerant- 114
(CClgCClFj) has a boiling point of 38.4° F under
atmospheric pressure. Evaporating and con-
densing pressures at standard ton conditions are
16.1 in. Hg and 22 psig, respectively. The com-
pressor displacement required is relatively low

REFRIGERANTS 297

for a low pressure refrigerant (19.59 cu ft/min/
ton at standard conditions) and the horsepower
required compares favorably with that required
by other common refrigerants.

Refrigerant- 1 14 is used with centrifugal com-
pressors in large commercial and industrial air
conditioning installations and for industrial
process water chilling down to —70° F. It is
also used with vane-type rotary compressors in
domestic refrigerators and in small drinking
water coolers.

Like Refrigerant-22, Refrigerant- 1 14 is oil
miscible under conditions found in the con-
densing section, but separates from oil in the
evaporator. However, because of the type of
equipment used with Refrigerant- 1 14 and the
conditions under which it is used, oil return is
not usually a problem.

Refrigerant-1 14 is a safe refrigerant. A halide
torch may be used for leak detection.
16-23. Straight Hydrocarbons. The straight
hydrocarbons are a group of fluids composed in
various proportions of the two elements hydro-
gen and carbon. Those having significance as
refrigerants are methane, ethane, butane, pro-
pane, ethylene, and isobutane. All are extremely
flammable and explosive. Too, since all act as
anesthetics in varying degrees, they are con-
sidered mildly toxic. Although none of these
compounds will absorb moisture to any appre-
ciable extent, all are extremely miscible with oil
under all conditions.

Although a few of the straight hydrocarbons
(butane, propane, and isobutane) have been
used in small quantities for domestic refrigera-
tion, their use is ordinarily limited to special
applications where an experienced attendent is
on duty. Ethane, methane, and ethylene are
employed to some extent in ultra-low tempera-
ture applications, usually in the lower stage of
two and three stage cascade systems. However,
even in these applications, it is likely that they
will be replaced in the future by Refrigerants- 13
and 14, the latter being used only in pilot plants
at the present time.

Leak detection is by soap solution only.
I6-Z4. Refrigerant-500. Refrigerant-500,
commonly known as Carrene 7*, is an azeo-
tropic mixturet of Refrigerant- 12 (73.8% by

* A proprietary refrigerant of the Carrier Corpor-
ation.

f An azeotropic mixture is a mixture of two or

weight) and Refrigerant- 152a (26.2%). It has a
boiling point at atmospheric pressure of —28° F.
Evaporator and condenser pressures at standard
ton conditions are 16.4 psig and 113.4 psig,
respectively. Although the horsepower require-
ments of Refrigerant-500 are approximately the
same as those for Refrigerants- 12 and 22, the
compressor displacement required is greater
than that required for Refrigerant-22, but some-
what less than that required for Refrigerant- 12.

The principal advantage of Refrigerant-500
lies in the fact that its substitution for Refrig-
erant- 12 results in an increase in compressor
capacity of approximately 18%. This makes it
possible to use the same direct connected com-
pressor (as in a hermetic motor-compressor
unit) on either 50 or 60 cycle power with little or
no change in the refrigerating capacity or in the
power requirements.

It will be shown in Chapter 21 that the speed
of an alternating current motor varies in direct
proportion to the cycle frequency. Therefore,
an electric motor operating on 50 cycle power
will have only five-sixths of the speed it has when
operating on 60 cycle power. For this reason,
the displacement of a direct connected com-
pressor is reduced approximately 18% when a
change is made from 60 to 50 cycle power.
Since the increase in capacity per unit of dis-
placement accruing from the substitution of
Refrigerant-500 for Refrigerant- 12 is almost
exactly equal to the loss of displacement suffered
when changing for 60 to 50 cycle power, the
same motor-compressor assembly is made
suitable for use with both frequencies by the
simple expedient of changing refrigerants.
16-25. Refrigerant Drying Agents. Refrig-
erant drying agents, called desiccants are fre-
quently employed in refrigerating systems to
remove moisture from the refrigerant. Some
of the most commonly used desiccants are silica
gel (silicon dioxide), activated alumina (alu-
minum oxide), and Drierite (anhydrous calcium
sulfate). Silica gel and activated alumina are
adsorption-type desiccants and are available in
granular form. Drierite is an absorption type
desiccant and is available in granular form and
in cast sticks.

more liquids, which, when mixed in precise pro-
portions, form a compound having a boiling
temperature which is independent of the boiling
temperatures of the individual liquids.

Refrigerant Flow
Controls

17-1. Types and Function. There are six
basic types of refrigerant flow controls: (1) the
hand expansion valve, (2) the automatic expan-
sion valve, (3) the thermostatic expansion valve,
(4) the capillary tube, (5) the low pressure float,
and (6) the high pressure float.

Regardless of type, the function of any refrig-
erant flow control is twofold: (1) to meter the
liquid refrigerant from the liquid line into the
evaporator at a rate commensurate with the rate
at which vaporization of the liquid is occurring
in the latter unit, and (2) to maintain a pressure
differential between the high and low pressure
sides of the system in order to permit the refrig-
erant to vaporize under the desired low pres-
sure in the evaporator while at the same time
condensing at a high pressure in the condenser.
17-2. Hand Expansion Valves. Hand ex-
pansion valves are hand-operated needle
valves (Fig. 17-1). The rate of liquid flow
through the valve depends on the pressure
differential across the valve orifice and on the
degree of valve opening, the latter being manu-
ally adjustable. Assuming that the pressure
differential across the valve remains the same,
the flow rate through a hand expansion valve
will remain constant at all times without regard
for either the evaporator pressure or the
evaporator loading.

The principal disadvantage of the hand
expansion valve is that it is unresponsive to
changes in the system load and therefore must be
manually readjusted each time the load on the
system changes in order to prevent either

starving or overfeeding of the evaporator,
depending upon the direction of the load shift.
Too, the valve must be opened and closed
manually each time the compressor is cycled on
and off.

Obviously the hand expansion valve is suitable
for use only on large systems where an operator
is on duty and where the load on the system is
relatively constant. When automatic control is
desired and/or when the system is subject to
frequent load fluctuations, some other type of
refrigerant flow control is required.

At the present time, the principal use of the
hand expansion valve is as an auxiliary refrig-
erant control installed in a by-pass line (Fig.
17-29). It is also frequently used to control the
flow rate through oil bleeder lines (Fig. 19-12).
17-3. Automatic Expansion Valves. A
schematic diagram of an automatic expansion
valve is shown in Fig. 17-2. The valve consists
mainly of a needle and seat, a pressure bellows
or diaphragm, and a spring, the tension of the
latter being variable by means of an adjusting
screw. A screen or strainer is usually installed
at the liquid inlet of the valve in order to pre-
vent the entrance of foreign materials which may
cause stoppage of the valve. The construction
of a typical automatic expansion valve is shown
in Fig. 17-3.

The automatic expansion valve functions to
maintain a constant pressure in the evaporator
by flooding more or less of the evaporator sur-
face in response to changes in the evaporator
load. The constant pressure characteristic of the
valve results from the interaction of two opposing
forces: (1) the evaporator pressure and (2) the
spring pressure. The evaporator pressure,
exerted on one side of the bellows or diaphragm,
acts to move the valve in a closing direction,
whereas the spring pressure, acting on the
opposite side of the bellows or diaphragm, acts
to move the valve in an opening direction. When
the compressor is running, the valve functions
to maintain the evaporator pressure in equi-
librium with the spring pressure.

As the name implies, the operation of the
valve is automatic and, once the tension of the
spring is adjusted for the desired evaporator
pressure, the valve will operate automatically
to regulate the flow of liquid refrigerant into the
evaporator so that the desired evaporator
pressure is maintained, regardless of evaporator

298

REFRIGERANT FLOW CONTROLS 299

loading. For example, assume that the tension
of the spring is adjusted to maintain a constant
pressure in the evaporator of lOpsig. There-
after, any time the evaporator pressure tends to
fall below lOpsig, the spring pressure will
exceed the evaporator pressure causing the valve
to move in the opening direction, thereby
increasing the flow of liquid to the evaporator
and flooding more of the evaporator surface.
As more of the evaporator surface becomes
effective, the rate of vaporization increases and

i" Flare

0.078" Orifice
f Flare

Fig. 17-1. Small capacity hand-expansion valve.
(Courtesy Mueller Brass Company.)

the evaporator pressure rises until equilibrium
is established with the spring pressure. Should
the evaporator pressure tend to rise above the
desired 10 psig, it will immediately override the
pressure of the spring and cause the valve to
move in the closing direction, thereby throttling
the flow of liquid into the evaporator and
reducing the amount of effective evaporator
surface. Naturally, this decreases the rate of
vaporization and lowers the evaporator pressure
until equilibrium is again established with the
spring pressure.

Adjusting
screw ~

Bellows or
diaphragm"" ?"^.

Needle

Strainer

Evaporator pressure

Fig. 17-2. Schematic diagram of automatic expansion
valve.

It is important to notice that the operating
characteristics of the automatic expansion valve
are such that the valve will close off tightly when
the compressor cycles off and remain closed
until the compressor cycles on again. As pre-
viously described, vaporization continues in the
evaporator for a short time after the com-
pressor cycles off and, since the resulting vapor
is not removed by the compressor, the pressure
in the evaporator rises. Hence, during the off
cycle, the evaporator pressure will always exceed
the spring pressure and the valve will be tightly
closed. When the compressor cycles on, the
evaporator pressure will be immediately reduced
below the spring pressure, at which time the
valve will open and admit sufficient liquid to the
evaporator to establish operating equilibrium
between the evaporator and spring pressures.

Fig. 17-3. Typical automatic expansion valve. (Cour-
tesy Controls Company of America.)

300 PRINCIPLES OF REFRIGERATION

Automatic

- expansion

valve

Liquid to
here

Liquid

from

receiver

(a)

Vapor!
compressor

Automatic

-expansion

valve

receiver

Liquid to
here

(b)

Vapor 1
compressor

Fig. 17-4. Operating characteristics of the auto-
matic expansion valve under varying load conditions.
(a) Heavy load conditions, (b) Minimum load
conditions.

The chief disadvantage of the automatic
expansion valve is its relatively poor efficiency
as compared to that of other refrigerant flow
controls. In view of the evaporator-compressor
relationship, it is evident that maintaining a
constant pressure in the evaporator requires that
the rate of vaporization in the evaporator be
kept constant. To accomplish this necessitates
severe throttling of the liquid in order to limit
the amount of effective evaporator surface when
the load on the evaporator is heavy and the heat
transfer capacity per unit of evaporator surface
is high (Fig. 17-4a). As the load on the evapora-
tor decreases and the heat transfer capacity per
unit of evaporator surface is reduced, more and
more of the evaporator surface must be flooded
with liquid if a constant rate of vaporization is
to be maintained (Fig. 17-46). As a matter of

fact, if the load on the evaporator is permitted to
fall below a certain level, the automatic expan-
sion valve, in an attempt to keep the evaporator
pressure up, will overfeed the evaporator to the
extent that liquid will enter the suction line and
be carried to the compressor where it may cause
serious damage. However, in a properly
designed system, overfeeding is not likely to
occur, since the thermostat will usually cycle the
compressor off before the space or product
temperature is reduced to a level such that the
load on the evaporator will fall below the
critical point.

Obviously, since it permits only a small
portion of the evaporator to be filled with liquid
during periods when the load on the system is
heavy, the constant pressure characteristic of the
automatic expansion valve severely limits the
capacity and efficiency of the refrigerating
system at a time when high capacity and high
efficiency are most desired. Too, because the
evaporator pressure is maintained constant
throughout the entire running cycle of the com-
pressor, the valve must be adjusted for a
pressure corresponding to the lowest evaporator
temperature required during the entire running
cycle (see Fig. 17-5). This results in a consider-
able loss in compressor capacity and efficiency,
since advantage cannot be taken of the higher
suction temperatures which would ordinarily
exist with a full-flooded evaporator during the
early part of the running cycle.

Another disadvantage of the automatic ex-
pansion valve, which can also be attributed to
its constant pressure characteristic, is that it
cannot be used in conjunction with a low
pressure motor control, since proper operation
of the latter part depends on a rather substantial
change in the evaporator pressure during the
running cycle, a condition which obviously

30
&|15

> (A
LU 0)

o. 5 «-0ff ->fr -0n->j.^0ff

0n-»f«— Off

5 10 15 20 25 30 35 40 45 50 55 60
Time in minutes

Fig. 17-5. Operating characteristics of the automatic
expansion valve.

REFRIGERANT FLOW CONTROLS 301

cannot be met when an automatic expansion
valve is used as the refrigerant flow control.

In view of its poor efficiency under heavy load
conditions, the automatic expansion valve is
best applied only to small equipment having
relatively constant loads, such as domestic refrig-
erators and freezers and small, retail ice-cream
storage cabinets. However, even in these appli-
cations the automatic expansion valve is seldom
used at the present time, having given way to
other types of refrigerant flow controls which
are more efficient and sometimes lower in
cost.

Some automatic expansion valves are now
being employed as "condenser by-pass valves."

evaporator pressure. In this respect, the con-
denser by-pass serves the same function as the
cylinder by-pass type of compressor capacity
control. * However, unlike the cylinder by-pass,
the condenser by-pass does not unload the
compressor in any way. Hence, with the con-
denser by-pass, there is no reduction in the work
of compression or in the power requirements of
the compressor. For this reason, the condenser
by-pass is not generally recommended as a
means of controlling the capacity of the
compressor.

Care should be taken to connect the by-pass
line to the condenser at a point low enough on
the condenser to insure that slightly "wet"

Fig. 17-6. Automatic expansion
valve employed as condenser
by-pass valve.

^> r

Thermostatic
• ^^"""expansion valve

Automatic expansion valve

adjusted for minimum desired

evaporator pressure

_ Condenser
"by-pass line

As such thay are installed in a by-pass line
between the condenser and the suction line (Fig.
17-6) where they serve to regulate the flow of hot
gas which is by-passed from the condenser
directly into the suction line in order to prevent
the evaporator pressure from dropping below a
predetermined desired minimum. In such cases,
the valve is set for the minimum desired eva-
porator pressure. As long as the pressure in the
evaporator remains above the desired minimum,
the valve will remain closed and no gas is by-
passed from the condenser into the suction line.
However, any time the evaporator tends to fall
below the desired minimum, the by-pass valve
opens and permits hot gas from the condenser
to pass directly into the suction line in an
amount just sufficient to maintain the minimum

vapor, rather than superheated vapor, is by-
passed to the suction line. Superheated vapor
dir